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• How-to delete a tree node using the context menu

  - by frank.nimphius

  Hierarchical trees in Oracle ADF make use of View Accessors, which means that only the top level node needs to be exposed as a View Object instance on the ADF Business Components Data Model. This also means that only the top level node has a representation in the PageDef file as a tree binding and iterator binding reference. Detail nodes are accessed through tree rule definitions that use the accessor mentioned above (or nested collections in the case of POJO or EJB business services). The tree component is configured for single node selection, which however can be declaratively changed for users to press the ctrl key and selecting multiple nodes. In the following, I explain how to create a context menu on the tree for users to delete the selected tree nodes. For this, the context menu item will access a managed bean, which then determines the selected node(s), the internal ADF node bindings and the rows they represent. As mentioned, the ADF Business Components Data Model only needs to expose the top level node data sources, which in this example is an instance of the Locations View Object. For the tree to work, you need to have associations defined between entities, which usually is done for you by Oracle JDeveloper if the database tables have foreign keys defined Note: As a general hint of best practices and to simplify your life: Make sure your database schema is well defined and designed before starting your development project. Don't treat the database as something organic that grows and changes with the requirements as you proceed in your project. Business service refactoring in response to database changes is possible, but should be treated as an exception, not the rule. Good database design is a necessity – even for application developers – and nothing evil. To create the tree component, expand the Data Controls panel and drag the View Object collection to the view. From the context menu, select the tree component entry and continue with defining the tree rules that make up the hierarchical structure. As you see, when pressing the green plus icon  in the Edit Tree Binding  dialog, the data structure, Locations -  Departments – Employees in my sample, shows without you having created a View Object instance for each of the nodes in the ADF Business Components Data Model. After you configured the tree structure in the Edit Tree Binding dialog, you press OK and the tree is created. Select the tree in the page editor and open the Structure Window (ctrl+shift+S). In the Structure window, expand the tree node to access the conextMenu facet. Use the right mouse button to insert a Popup  into the facet. Repeat the same steps to insert a Menu and a Menu Item into the Popup you created. The Menu item text should be changed to something meaningful like "Delete". Note that the custom menu item later is added to the context menu together with the default context menu options like expand and expand all. To define the action that is executed when the menu item is clicked on, you select the Action Listener property in the Property Inspector and click the arrow icon followed by the Edit menu option. Create or select a managed bean and define a method name for the action handler. Next, select the tree component and browse to its binding property in the Property Inspector. Again, use the arrow icon | Edit option to create a component binding in the same managed bean that has the action listener defined. The tree handle is used in the action listener code, which is shown below: public void onTreeNodeDelete(ActionEvent actionEvent) {   //access the tree from the JSF component reference created   //using the af:tree "binding" property. The "binding" property   //creates a pair of set/get methods to access the RichTree instance   RichTree tree = this.getTreeHandler();   //get the list of selected row keys   RowKeySet rks = tree.getSelectedRowKeys();   //access the iterator to loop over selected nodes   Iterator rksIterator = rks.iterator();          //The CollectionModel represents the tree model and is   //accessed from the tree "value" property   CollectionModel model = (CollectionModel) tree.getValue();   //The CollectionModel is a wrapper for the ADF tree binding   //class, which is JUCtrlHierBinding   JUCtrlHierBinding treeBinding =                  (JUCtrlHierBinding) model.getWrappedData();          //loop over the selected nodes and delete the rows they   //represent   while(rksIterator.hasNext()){     List nodeKey = (List) rksIterator.next();     //find the ADF node binding using the node key     JUCtrlHierNodeBinding node =                       treeBinding.findNodeByKeyPath(nodeKey);     //delete the row.     Row rw = node.getRow();       rw.remove();   }          //only refresh the tree if tree nodes have been selected   if(rks.size() > 0){     AdfFacesContext adfFacesContext =                          AdfFacesContext.getCurrentInstance();     adfFacesContext.addPartialTarget(tree);   } } Note: To enable multi node selection for a tree, select the tree and change the row selection setting from "single" to "multiple". Note: a fully pictured version of this post will become available at the end of the month in a PDF summary on ADF Code Corner : http://www.oracle.com/technetwork/developer-tools/adf/learnmore/index-101235.html 

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• Event Driven Behavior Tree: deterministic traversal order with parallel

  - by Heisenbug

  I've studied several articles and listen some talks about behavior trees (mostly the resources available on AIGameDev by Alex J. Champandard). I'm particularly interested on event driven behavior trees, but I have still some doubts on how to implement them correctly using a scheduler. Just a quick recap: Standard Behavior Tree Each execution tick the tree is traversed from the root in depth-first order The execution order is implicitly expressed by the tree structure. So in the case of behaviors parented to a parallel node, even if both children are executed during the same traversing, the first leaf is always evaluated first. Event Driven BT During the first traversal the nodes (tasks) are enqueued using a scheduler which is responsible for updating only running ones every update The first traversal implicitly produce a depth-first ordered queue in the scheduler Non leaf nodes stays suspended mostly of the time. When a leaf node terminate(either with success or fail status) the parent (observer) is waked up allowing the tree traversing to continue and new tasks will be enqueued in the scheduler Without parallel nodes in the tree there will be up to 1 task running in the scheduler Without parallel nodes, the tasks in the queue(excluding dynamic priority implementation) will be always ordered in a depth-first order (is this right?) Now, from what is my understanding of a possible implementation, there are 2 requirements I think must be respected(I'm not sure though): Now, some requirements I think needs to be guaranteed by a correct implementation are: The result of the traversing should be independent from which implementation strategy is used. The traversing result must be deterministic. I'm struggling trying to guarantee both in the case of parallel nodes. Here's an example: Parallel_1 -->Sequence_1 ---->leaf_A ---->leaf_B -->leaf_C Considering a FIFO policy of the scheduler, before leaf_A node terminates the tasks in the scheduler are: P1(suspended),S1(suspended),leaf_A(running),leaf_C(running) When leaf_A terminate leaf_B will be scheduled (at the end of the queue), so the queue will become: P1(suspended),S1(suspended),leaf_C(running),leaf_B(running) In this case leaf_B will be executed after leaf_C at every update, meanwhile with a non event-driven traversing from the root node, the leaf_B will always be evaluated before leaf_A. So I have a couple of question: do I have understand correctly how event driven BT work? How can I guarantee the depth first order is respected with such an implementation? is this a common issue or am I missing something?

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• What is the kd tree intersection logic?

  - by bobobobo

  I'm trying to figure out how to implement a KD tree. On page 322 of "Real time collision detection" by Ericson The text section is included below in case Google book preview doesn't let you see it the time you click the link text section Relevant section: The basic idea behind intersecting a ray or directed line segment with a k-d tree is straightforward. The line is intersected against the node's splitting plane, and the t value of intersection is computed. If t is within the interval of the line, 0 <= t <= tmax, the line straddles the plane and both children of the tree are recursively descended. If not, only the side containing the segment origin is recursively visited. So here's what I have: (open image in new tab if you can't see the lettering) The logical tree Here the orange ray is going thru the 3d scene. The x's represent intersection with a plane. From the LEFT, the ray hits: The front face of the scene's enclosing cube, The (1) splitting plane The (2.2) splitting plane The right side of the scene's enclosing cube But here's what would happen, naively following Ericson's basic description above: Test against splitting plane (1). Ray hits splitting plane (1), so left and right children of splitting plane (1) are included in next test. Test against splitting plane (2.1). Ray actually hits that plane, (way off to the right) so both children are included in next level of tests. (This is counter-intuitive - shouldn't only the bottom node be included in subsequent tests) Can some one describe what happens when the orange ray goes through the scene correctly?

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• What is a B-tree page

  - by Phenom

  I think I know what a B-tree is but what is a B-tree page?

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• Algorithm for parsing a flat tree into a non-flat tree

  - by Chad Johnson

  I have the following flat tree: id name parent_id is_directory =========================================================== 50 app 0 1 31 controllers 50 1 11 application_controller.rb 31 0 46 models 50 1 12 test_controller.rb 31 0 31 test.rb 46 0 and I am trying to figure out an algorithm for getting this into the following tree structuree: [{ id: 50, name: app, is_directory: true children: [{ id: 31, name: controllers, is_directory: true, children: [{ id: 11, name: application_controller.rb is_directory: false },{ id: 12, name: test_controller.rb, is_directory: false }], },{ id: 46, name: models, is_directory: true, children: [{ id: 31, name: test.rb, is_directory: false }] }] }] Can someone point me in the right direction? I'm looking for steps (eg. build an associative array; loop through the array looking for x; etc.).

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• Implementation of Race Game Tree

  - by Mert Toka

  I build a racing game right in OpenGL using Glut, and I'm a bit lost in all the details. First of all, any suggestions as a road map would be more than great. So far what I thought is this: Tree implementation for transformations. Simulated dynamics.(*) Octree implementation for collusion detection. Actual collusion detection.(*) Modelling in Maya and export them as .OBJs. Polishing the game with GLSL or something like that for graphics quality. (*): I am not sure the order of these two. So I started with the simulated dynamics without tree, and it turned out to be a huge chaos for me. Is there any way you can think of such that could help me to build such tree to use in racing game? I thought something like this but I have no idea how to implement it. Reds are static, yellows are dynamic nodes

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• Programmatically disclosing a node in af:tree and af:treeTable

  - by Frank Nimphius

  A common developer requirement when working with af:tree or af:treeTable components is to programmatically disclose (expand) a specific node in the tree. If the node to disclose is not a top level node, like a location in a LocationsView -> DepartmentsView -> EmployeesView hierarchy, you need to also disclose the node's parent node hierarchy for application users to see the fully expanded tree node structure. Working on ADF Code Corner sample #101, I wrote the following code lines that show a generic option for disclosing a tree node starting from a handle to the node to disclose. The use case in ADF Coder Corner sample #101 is a drag and drop operation from a table component to a tree to relocate employees to a new department. The tree node that receives the drop is a department node contained in a location. In theory the location could be part of a country and so on to indicate the depth the tree may have. Based on this structure, the code below provides a generic solution to parse the current node parent nodes and its child nodes. The drop event provided a rowKey for the tree node that received the drop. Like in af:table, the tree row key is not of type oracle.jbo.domain.Key but an implementation of java.util.List that contains the row keys. The JUCtrlHierBinding class in the ADF Binding layer that represents the ADF tree binding at runtime provides a method named findNodeByKeyPath that allows you to get a handle to the JUCtrlHierNodeBinding instance that represents a tree node in the binding layer. CollectionModel model = (CollectionModel) your_af_tree_reference.getValue(); JUCtrlHierBinding treeBinding = (JUCtrlHierBinding ) model.getWrappedData(); JUCtrlHierNodeBinding treeDropNode = treeBinding.findNodeByKeyPath(dropRowKey); To disclose the tree node, you need to create a RowKeySet, which you do using the RowKeySetImpl class. Because the RowKeySet replaces any existing row key set in the tree, all other nodes are automatically closed. RowKeySetImpl rksImpl = new RowKeySetImpl(); //the first key to add is the node that received the drop //operation (departments).            rksImpl.add(dropRowKey);    Similar, from the tree binding, the root node can be obtained. The root node is the end of all parent node iteration and therefore important. JUCtrlHierNodeBinding rootNode = treeBinding.getRootNodeBinding(); The following code obtains a reference to the hierarchy of parent nodes until the root node is found. JUCtrlHierNodeBinding dropNodeParent = treeDropNode.getParent(); //walk up the tree to expand all parent nodes while(dropNodeParent != null && dropNodeParent != rootNode){    //add the node's keyPath (remember its a List) to the row key set    rksImpl.add(dropNodeParent.getKeyPath());      dropNodeParent = dropNodeParent.getParent(); } Next, you disclose the drop node immediate child nodes as otherwise all you see is the department node. Its not quite exactly "dinner for one", but the procedure is very similar to the one handling the parent node keys ArrayList<JUCtrlHierNodeBinding> childList = (ArrayList<JUCtrlHierNodeBinding>) treeDropNode.getChildren();                     for(JUCtrlHierNodeBinding nb : childList){   rksImpl.add(nb.getKeyPath()); } Next, the row key set is defined as the disclosed row keys on the tree so when you refresh (PPR) the tree, the new disclosed state shows tree.setDisclosedRowKeys(rksImpl); AdfFacesContext.getCurrentInstance().addPartialTarget(tree.getParent()); The refresh in my use case is on the tree parent component (a layout container), which usually shows the best effect for refreshing the tree component. 

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• Find the minimum gap between two numbers in an AVL tree

  - by user1656647

  I have a data structures homework, that in addition to the regular AVL tree functions, I have to add a function that returns the minimum gap between any two numbers in the AVL tree (the nodes in the AVL actually represent numbers.) Lets say we have the numbers (as nodes) 1 5 12 20 23 21 in the AVL tree, the function should return the minimum gap between any two numbers. In this situation it should return "1" which is |20-21| or |21-20|. It should be done in O(1). Tried to think alot about it, and I know there is a trick but just couldn't find it, I have spent hours on this. There was another task which is to find the maximum gap, which is easy, it is the difference between the minimal and maximal number.

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• Can a binary tree or tree be always represented in a Database as 1 table and self-referencing?

  - by Jian Lin

  I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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• Can a binary tree or tree be always represented in a Database table as 1 table and self-referencing?

  - by Jian Lin

  I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person under him can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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• Inorder tree traversal in binary tree in C

  - by srk

  In the below code, I'am creating a binary tree using insert function and trying to display the inserted elements using inorder function which follows the logic of In-order traversal.When I run it, numbers are getting inserted but when I try the inorder function( input 3), the program continues for next input without displaying anything. I guess there might be a logical error.Please help me clear it. Thanks in advance... #include<stdio.h> #include<stdlib.h> int i; typedef struct ll { int data; struct ll *left; struct ll *right; } node; node *root1=NULL; // the root node void insert(node *root,int n) { if(root==NULL) //for the first(root) node { root=(node *)malloc(sizeof(node)); root->data=n; root->right=NULL; root->left=NULL; } else { if(n<(root->data)) { root->left=(node *)malloc(sizeof(node)); insert(root->left,n); } else if(n>(root->data)) { root->right=(node *)malloc(sizeof(node)); insert(root->right,n); } else { root->data=n; } } } void inorder(node *root) { if(root!=NULL) { inorder(root->left); printf("%d ",root->data); inorder(root->right); } } main() { int n,choice=1; while(choice!=0) { printf("Enter choice--- 1 for insert, 3 for inorder and 0 for exit\n"); scanf("%d",&choice); switch(choice) { case 1: printf("Enter number to be inserted\n"); scanf("%d",&n); insert(root1,n); break; case 3: inorder(root1); break; default: break; } } }

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• AVL tree in C language

  - by I_S_W

  Hey all; i am currently doing a project that requires the use of AVL trees , the insert function i wrote for the avl does not seem to be working , it works for 3 or 4 nodes at maximum ; i would really appreciate your help The attempt is below enter code here Tree insert(Tree t,char name[80],int num) { if(t==NULL) { t=(Tree)malloc(sizeof(struct node)); if(t!=NULL) { strcpy(t->name,name); t->num=num; t->left=NULL; t->right=NULL; t->height=0; } } else if(strcmp(name,t->name)<0) { t->left=insert(t->left,name,num); if((height(t->left)-height(t->right))==2) if(strcmp(name,t->left->name)<0) { t=s_rotate_left(t);} else{ t=d_rotate_left(t);} } else if(strcmp(name,t-name)0) { t-right=insert(t-right,name,num); if((height(t-right)-height(t-left))==2) if(strcmp(name,t-right-name)0){ t=s_rotate_right(t); } else{ t=d_rotate_right(t);} } t-height=max(height(t-left),height(t-right))+1; return t; }

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• New to AVL tree implementation.

  - by nn

  I am writing a sliding window compression algorithm (LZ77) that searches for phrases in a "moving" dictionary. So far I have written a BST where each node is stored in an array and it's index in the array is also the value of the starting position in the window itself. I am now looking at transforming the BST to an AVL tree. I am a little confused at the sample implementations I have seen. Some only appear to store the balance factors whereas others store the height of each tree. Are there any performance advantage/disadvantages of storing the height and/or balance factor for each node? Apologies if this is a very simple question, but I'm still not visualizing how I want to restructure my BST to implement height balancing. Thanks.

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• Convert a binary tree to linked list, breadth first, constant storage/destructive

  - by Merlyn Morgan-Graham

  This is not homework, and I don't need to answer it, but now I have become obsessed :) The problem is: Design an algorithm to destructively flatten a binary tree to a linked list, breadth-first. Okay, easy enough. Just build a queue, and do what you have to. That was the warm-up. Now, implement it with constant storage (recursion, if you can figure out an answer using it, is logarithmic storage, not constant). I found a solution to this problem on the Internet about a year back, but now I've forgotten it, and I want to know :) The trick, as far as I remember, involved using the tree to implement the queue, taking advantage of the destructive nature of the algorithm. When you are linking the list, you are also pushing an item into the queue. Each time I try to solve this, I lose nodes (such as each time I link the next node/add to the queue), I require extra storage, or I can't figure out the convoluted method I need to get back to a node that has the pointer I need. Even the link to that original article/post would be useful to me :) Google is giving me no joy. Edit: Jérémie pointed out that there is a fairly simple (and well known answer) if you have a parent pointer. While I now think he is correct about the original solution containing a parent pointer, I really wanted to solve the problem without it :) The refined requirements use this definition for the node: struct tree_node { int value; tree_node* left; tree_node* right; };

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• How to functionally generate a tree breadth-first. (With Haskell)

  - by Dennetik

  Say I have the following Haskell tree type, where "State" is a simple wrapper: data Tree a = Branch (State a) [Tree a] | Leaf (State a) deriving (Eq, Show) I also have a function "expand :: Tree a - Tree a" which takes a leaf node, and expands it into a branch, or takes a branch and returns it unaltered. This tree type represents an N-ary search-tree. Searching depth-first is a waste, as the search-space is obviously infinite, as I can easily keep on expanding the search-space with the use of expand on all the tree's leaf nodes, and the chances of accidentally missing the goal-state is huge... thus the only solution is a breadth-first search, implemented pretty decent over here, which will find the solution if it's there. What I want to generate, though, is the tree traversed up to finding the solution. This is a problem because I only know how to do this depth-first, which could be done by simply called the "expand" function again and again upon the first child node... until a goal-state is found. (This would really not generate anything other then a really uncomfortable list.) Could anyone give me any hints on how to do this (or an entire algorithm), or a verdict on whether or not it's possible with a decent complexity? (Or any sources on this, because I found rather few.)

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• The "Ruby" way of doing an n-ary tree

  - by sardaukar

  I'm writing a Ruby script and would like to use a n-ary tree data structure. Is there a good implementation that is available as source code? Thanks.

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• Spanning-tree setup with incompatible switches

  - by wfaulk

  I have a set of eight HP ProCurve 2910al-48G Ethernet switches at my datacenter that are set up in a star topology with no physical loops. I want to partially mesh the switches for redundancy and manage the loops with a spanning-tree protocol. However, our connection to the datacenter is provided by two uplinks, each to a Cisco 3750. The datacenter's switches are handling the redundant connection using PVST spanning-tree, which is a Cisco-proprietary spanning-tree implementation that my HP switches do not support. It appears that my switches are not participating in the datacenter's spanning-tree domain, but are blindly passing the BPDUs between the two switchports on my side, which enables the datacenter's switches to recognize the loop and put one of the uplinks into the Blocking state. This is somewhat supposition, but I can confirm that, while my switches say that both of the uplink ports are forwarding, only one is passing any real quantity of data. (I am assuming that I cannot get the datacenter to move away from PVST. I don't know that I'd want them to make that significant of a change anyway.) The datacenter has also sent me this output from their switches (which I have expurgated of any identifiable info): 3750G-1#sh spanning-tree vlan nnn VLAN0nnn Spanning tree enabled protocol ieee Root ID Priority 10 Address 00d0.0114.xxxx Cost 4 Port 5 (GigabitEthernet1/0/5) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32mmm (priority 32768 sys-id-ext nnn) Address 0018.73d3.yyyy Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Gi1/0/5 Root FWD 4 128.5 P2p Gi1/0/6 Altn BLK 4 128.6 P2p Gi1/0/8 Altn BLK 4 128.8 P2p and: 3750G-2#sh spanning-tree vlan nnn VLAN0nnn Spanning tree enabled protocol ieee Root ID Priority 10 Address 00d0.0114.xxxx Cost 4 Port 6 (GigabitEthernet1/0/6) Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Bridge ID Priority 32mmm (priority 32768 sys-id-ext nnn) Address 000f.f71e.zzzz Hello Time 2 sec Max Age 20 sec Forward Delay 15 sec Aging Time 300 sec Interface Role Sts Cost Prio.Nbr Type ------------------- ---- --- --------- -------- -------------------------------- Gi1/0/1 Desg FWD 4 128.1 P2p Gi1/0/5 Altn BLK 4 128.5 P2p Gi1/0/6 Root FWD 4 128.6 P2p Gi1/0/8 Desg FWD 4 128.8 P2p The uplinks to my switches are on Gi1/0/8 on both of their switches. The uplink ports are configured with a single tagged VLAN. I am also using a number of other tagged VLANs in my switch infrastructure. And, to be clear, I am passing the tagged VLAN I'm receiving from the datacenter to other ports on other switches in my infrastructure. My question is: how do I configure my switches so that I can use a spanning tree protocol inside my switch infrastructure without breaking the datacenter's spanning tree that I cannot participate in?

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• ASP.NET, URL Rewriting and threads on Linux?

  - by acidzombie24

  Right now converting my normal sql(ite) syntax into compatible MS Server syntax is driving me mad. My code seems to be mono compatible judging by MoMA. However in global.asax i start several threads in Application_Start to use as a service and excessive url rewrites in Application_BeginRequest (everything is rewriting into default.aspx which calls C# code/class baded on the url). The urls do not have extensions or my have false extentions ie mysite.com/username/page/title.lol - id I need it all to go to BeginRequest. I also need files to be downloadable. Usually the path will be rewritten as from /file/user/filename to /file/u/s/user/file. Would this be supported on linux+mono+a good webserver?

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• What is the difference between an Abstract Syntax Tree and a Concrete Syntax Tree?

  - by Jason Baker

  I've been reading a bit about how interpreters/compilers work, and one area where I'm getting confused is the difference between an AST and a CST. My understanding is that the parser makes a CST, hands it to the semantic analyzer which turns it into an AST. However, my understanding is that the semantic analyzer simply ensures that rules are followed. I don't really understand why it would actually make any changes to make it abstract rather than concrete. Is there something that I'm missing about the semantic analyzer, or is the difference between an AST and CST somewhat artificial?

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• Token Suffix Tree Tutorial

  - by Bart J

  Can someone please point to tutorials on - "Token Suffix Trees". Google, gives only links to research papers that are already using them! :( Thanks in advance.

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• Deletion procedure for a Binary Search Tree

  - by Metz

  Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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• Return parent of node in Binary Tree

  - by user188995

  I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs. //Assume that nodes are represented by numbers from 1...n where 1=root and even //nos.=left child and odd nos=right child. public int parent(Node node){ if (node % 2 == 0){ if (root.left==node) return root; else return parent(root.left); } //same case for right } But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right. If it's the child or if the node that was queried else, recurses it with the child.

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• Problems in Binary Search Tree

  - by user2782324

  This is my first ever trial at implementing the BST, and I am unable to get it done. Please help The problem is that When I delete the node if the node is in the right subtree from the root or if its a right child in the left subtree, then it works fine. But if the node is in the left subtree from root and its any left child, then it does not get deleted. Can someone show me what mistake am I doing?? the markedNode here gets allocated to the parent node of the node to be deleted. the minValueNode here gets allocated to a node whose left value child is the smallest value and it will be used to replace the value to be deleted. package DataStructures; class Node { int value; Node rightNode; Node leftNode; } class BST { Node rootOfTree = null; public void insertintoBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { Node newNode = new Node(); newNode.value = value; rootOfTree = newNode; newNode.rightNode = null; newNode.leftNode = null; } else { while (true) { if (value >= markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.rightNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { Node newNode = new Node(); newNode.value = value; markedNode.leftNode = newNode; newNode.rightNode = null; newNode.leftNode = null; break; } } } } } public void searchBST(int value) { Node markedNode = rootOfTree; if (rootOfTree == null) { System.out.println("Element Not Found"); } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { markedNode = markedNode.rightNode; } else { System.out.println("Element Not Found"); break; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { markedNode = markedNode.leftNode; } else { System.out.println("Element Not Found"); break; } } if (value == markedNode.value) { System.out.println("Element Found"); break; } } } } public void deleteFromBST(int value) { Node markedNode = rootOfTree; Node minValueNode = null; if (rootOfTree == null) { System.out.println("Element Not Found"); return; } if (rootOfTree.value == value) { if (rootOfTree.leftNode == null && rootOfTree.rightNode == null) { rootOfTree = null; return; } else if (rootOfTree.leftNode == null ^ rootOfTree.rightNode == null) { if (rootOfTree.rightNode != null) { rootOfTree = rootOfTree.rightNode; return; } else { rootOfTree = rootOfTree.leftNode; return; } } else { minValueNode = rootOfTree.rightNode; if (minValueNode.leftNode == null) { rootOfTree.rightNode.leftNode = rootOfTree.leftNode; rootOfTree = rootOfTree.rightNode; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node rootOfTree.value = minValueNode.leftNode.value; // The value has been swapped if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } else { while (true) { if (value > markedNode.value) { if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { break; } else { markedNode = markedNode.rightNode; } } else { System.out.println("Element Not Found"); return; } } if (value < markedNode.value) { if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { break; } else { markedNode = markedNode.leftNode; } } else { System.out.println("Element Not Found"); return; } } } // Parent of the required element found // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { if (markedNode.rightNode.rightNode == null && markedNode.rightNode.leftNode == null) { markedNode.rightNode = null; return; } else if (markedNode.rightNode.rightNode == null ^ markedNode.rightNode.leftNode == null) { if (markedNode.rightNode.rightNode != null) { markedNode.rightNode = markedNode.rightNode.rightNode; return; } else { markedNode.rightNode = markedNode.rightNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.rightNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////// if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { if (markedNode.leftNode.rightNode == null && markedNode.leftNode.leftNode == null) { markedNode.leftNode = null; return; } else if (markedNode.leftNode.rightNode == null ^ markedNode.leftNode.leftNode == null) { if (markedNode.leftNode.rightNode != null) { markedNode.leftNode = markedNode.leftNode.rightNode; return; } else { markedNode.leftNode = markedNode.leftNode.leftNode; return; } } else { if (markedNode.rightNode.value == value) { minValueNode = markedNode.rightNode.rightNode; } else { minValueNode = markedNode.leftNode.rightNode; } if (minValueNode.leftNode == null) { // MinNode has no left value markedNode.leftNode = minValueNode; return; } else { while (true) { if (minValueNode.leftNode.leftNode != null) { minValueNode = minValueNode.leftNode; } else { break; } } // Minvalue to the left of minvalue node if (markedNode.leftNode != null) { if (markedNode.leftNode.value == value) { markedNode.leftNode.value = minValueNode.leftNode.value; } } if (markedNode.rightNode != null) { if (markedNode.rightNode.value == value) { markedNode.rightNode.value = minValueNode.leftNode.value; } } // MarkedNode exchanged if (minValueNode.leftNode.leftNode == null && minValueNode.leftNode.rightNode == null) { minValueNode.leftNode = null; } else { if (minValueNode.leftNode.leftNode != null) { minValueNode.leftNode = minValueNode.leftNode.leftNode; } else { minValueNode.leftNode = minValueNode.leftNode.rightNode; } // Minvalue deleted } } } } // //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// } } } } } } public class BSTImplementation { public static void main(String[] args) { BST newBst = new BST(); newBst.insertintoBST(19); newBst.insertintoBST(13); newBst.insertintoBST(10); newBst.insertintoBST(20); newBst.insertintoBST(5); newBst.insertintoBST(23); newBst.insertintoBST(28); newBst.insertintoBST(16); newBst.insertintoBST(27); newBst.insertintoBST(9); newBst.insertintoBST(4); newBst.insertintoBST(22); newBst.insertintoBST(17); newBst.insertintoBST(30); newBst.insertintoBST(40); newBst.deleteFromBST(5); newBst.deleteFromBST(4); newBst.deleteFromBST(9); newBst.deleteFromBST(10); newBst.deleteFromBST(13); newBst.deleteFromBST(16); newBst.deleteFromBST(17); newBst.searchBST(5); newBst.searchBST(4); newBst.searchBST(9); newBst.searchBST(10); newBst.searchBST(13); newBst.searchBST(16); newBst.searchBST(17); System.out.println(); newBst.deleteFromBST(20); newBst.deleteFromBST(23); newBst.deleteFromBST(27); newBst.deleteFromBST(28); newBst.deleteFromBST(30); newBst.deleteFromBST(40); newBst.searchBST(20); newBst.searchBST(23); newBst.searchBST(27); newBst.searchBST(28); newBst.searchBST(30); newBst.searchBST(40); } }

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• Problem with building tree bottom up

  - by Esmond

  Hi, I have problems building a binary tree from the bottom up. THe input of the tree would be internal nodes of the trees with the children of this node being the leaves of the eventual tree. So initially if the tree is empty the root would be the first internal node. Afterwards, The next internal node to be added would be the new root(NR), with the old root(OR) being one of the child of NR. And so on. The problem i have is that whenever i add a NR, the children of the OR seems to be lost when i do a inOrder traversal. This is proven to be the case when i do a getSize() call which returns the same number of nodes before and after addNode(Tree,Node) Any help with resolving this problem is appreciated edited with the inclusion of node class code. both tree and node classes have the addChild methods because i'm not very sure where to put them for it to be appropriated. any comments on this would be appreciated too. The code is as follows: import java.util.*; public class Tree { Node root; int size; public Tree() { root = null; } public Tree(Node root) { this.root = root; } public static void setChild(Node parent, Node child, double weight) throws ItemNotFoundException { if (parent.child1 != null && parent.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (parent.child1 != null) { parent.child2 = child; child.parent = parent; parent.c2Weight = weight; } else { parent.child1 = child; child.parent = parent; parent.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public static Tree addNode(Tree tree, Node node) throws ItemNotFoundException { Tree tree1; if (tree.root == null) { tree.root = node; } else if (tree.root.getSeq().equals(node.getChild1().getSeq()) || tree.root.getSeq().equals(node.getChild2().getSeq())) { Node oldRoot = tree.root; oldRoot.setParent(node); tree.root = node; } else { //form a disjoint tree and merge the 2 trees tree1 = new Tree(node); tree = mergeTree(tree, tree1); } System.out.print("addNode2 = "); if(tree.root != null ) { Tree.inOrder(tree.root); } System.out.println(); return tree; } public static Tree mergeTree(Tree tree, Tree tree1) { String root = "root"; Node node = new Node(root); tree.root.setParent(node); tree1.root.setParent(node); tree.root = node; return tree; } public static int getSize(Node root) { if (root != null) { return 1 + getSize(root.child1) + getSize(root.child2); } else { return 0; } } public static boolean isEmpty(Tree Tree) { return Tree.root == null; } public static void inOrder(Node root) { if (root != null) { inOrder(root.child1); System.out.print(root.sequence + " "); inOrder(root.child2); } } } public class Node { Node child1; Node child2; Node parent; double c1Weight; double c2Weight; String sequence; boolean isInternal; public Node(String seq) { sequence = seq; child1 = null; c1Weight = 0; child2 = null; c2Weight = 0; parent = null; isInternal = false; } public boolean hasChild() { if (this.child1 == null && this.child2 == null) { this.isInternal = false; return isInternal; } else { this.isInternal = true; return isInternal; } } public String getSeq() throws ItemNotFoundException { if (this.sequence == null) { throw new ItemNotFoundException("No such node"); } else { return this.sequence; } } public void setChild(Node child, double weight) throws ItemNotFoundException { if (this.child1 != null && this.child2 != null) { throw new ItemNotFoundException("This Node already has 2 children"); } else if (this.child1 != null) { this.child2 = child; this.c2Weight = weight; } else { this.child1 = child; this.c1Weight = weight; } } public static void setChild1(Node parent, Node child) { parent.child1 = child; child.parent = parent; } public static void setChild2(Node parent, Node child) { parent.child2 = child; child.parent = parent; } public void setParent(Node parent){ this.parent = parent; } public Node getParent() throws ItemNotFoundException { if (this.parent == null) { throw new ItemNotFoundException("This Node has no parent"); } else { return this.parent; } } public Node getChild1() throws ItemNotFoundException { if (this.child1 == null) { throw new ItemNotFoundException("There is no child1"); } else { return this.child1; } } public Node getChild2() throws ItemNotFoundException { if (this.child2 == null) { throw new ItemNotFoundException("There is no child2"); } else { return this.child2; } } }

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• Behaviour tree code example?

  - by jokoon

  http://altdevblogaday.org/2011/02/24/introduction-to-behavior-trees/ Obviously the most interesting article I found on this website. What do you think about it ? It lacks some code example, don't you know any ? I also read that state machines are not very flexible compared to behaviour trees... On top of that I'm not sure if there is a true link between state machines and the state pattern... is there ?

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