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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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• C++ find largest BST in a binary tree

  - by fonjibe

  what is your approach to have the largest BST in a binary tree? I refer to this post where a very good implementation for finding if a tree is BST or not is bool isBinarySearchTree(BinaryTree * n, int min=std::numeric_limits<int>::min(), int max=std::numeric_limits<int>::max()) { return !n || (min < n->value && n->value < max && isBinarySearchTree(n->l, min, n->value) && isBinarySearchTree(n->r, n->value, max)); } It is quite easy to implement a solution to find whether a tree contains a binary search tree. i think that the following method makes it: bool includeSomeBST(BinaryTree* n) { if(!isBinarySearchTree(n)) { if(!isBinarySearchTree(n->left)) return isBinarySearchTree(n->right); } else return true; else return true; } but what if i want the largest BST? this is my first idea, BinaryTree largestBST(BinaryTree* n) { if(isBinarySearchTree(n)) return n; if(!isBinarySearchTree(n->left)) { if(!isBinarySearchTree(n->right)) if(includeSomeBST(n->right)) return largestBST(n->right); else if(includeSomeBST(n->left)) return largestBST(n->left); else return NULL; else return n->right; } else return n->left; } but its not telling the largest actually. i struggle to make the comparison. how should it take place? thanks

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• Find kth smallest element in a binary search tree in Optimum way

  - by Bragaadeesh

  Hi, I need to find the kth smallest element in the binary search tree without using any static/global variable. How to achieve it efficiently? The solution that I have in my mind is doing the operation in O(n), the worst case since I am planning to do an inorder traversal of the entire tree. But deep down I feel that I am not using the BST property here. Is my assumptive solution correct or is there a better one available ?

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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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• C++ design question on traversing binary trees

  - by user231536

  I have a binary tree T which I would like to copy to another tree. Suppose I have a visit method that gets evaluated at every node: struct visit { virtual void operator() (node* n)=0; }; and I have a visitor algorithm void visitor(node* t, visit& v) { //do a preorder traversal using stack or recursion if (!t) return; v(t); visitor(t->left, v); visitor(t->right, v); } I have 2 questions: I settled on using the functor based approach because I see that boost graph does this (vertex visitors). Also I tend to repeat the same code to traverse the tree and do different things at each node. Is this a good design to get rid of duplicated code? What other alternative designs are there? How do I use this to create a new binary tree from an existing one? I can keep a stack on the visit functor if I want, but it gets tied to the algorithm in visitor. How would I incorporate postorder traversals here ? Another functor class?

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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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• 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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• 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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• How to add add node in binary tree

  - by m.qayyum

  (5)Root (3)-------^--------(7) (2)---^----(5) ^-----(8) i want to add node with data 5 in this binray tree....Please help.....i really stuck in it

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• represent binary search trees in python

  - by Bunny Rabbit

  how do i represent binary search trees in python?

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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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• Do dconf use EXI binary XML?

  - by Hibou57

  A question came to my mind reading an answer to the question What are the differences between gconf and dconf?. In an reply to the above question, Oli said: Binary read access is far faster than parsing XML. However, there exist a W3C recommendation for binary XML, since 2010: Efficient XML Interchange (EXI) Format 1.0. Is this what dconf uses? If Yes, where is it confirmed? If No, was there some investigations toward it at some time, and what was the conclusions? Thanks for any tracks, I'm curious to know.

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• Binary on the Coat of Arms of the Governor General of Canada

  - by user132636

  Can you help me further this investigation? Here is about 10% of the work I have done on it. I present it only to see if there are any truly curious people among you. I made a video a few weeks ago showing some strange things about the Governor General's Coat of Arms and the binary on it. Today, I noticed something kinda cool and thought I would share. Here is the binary as it appears on the COA: 110010111001001010100100111010011 As DEC: 6830770643 (this is easily found on the web) Take a close look at that number. What do you notice about it? It has a few interesting features, but here is the one no one has pointed out... Split it down the middle and you have 68307 70643. The first digit is double the value of the last digit. The second digit is double the second last digit. The third digit is half of the third to last digit. And the middle ones are even or neutral. At first, I thought of it as energy. ++-nnnn+-- But actually you can create something else with it using the values. 221000211. See how that works. You may be asking why that is significant. Bare with me. I know 99% are rolling their eyes. 221000211 as base3 gives you this as binary: 100011101000111 100011101000111 as HEX is 4747, which converts to "GG". Initials of Governor General. GG.ca is his website. When you convert to base 33 (there are 33 digits in the original code) you get "GOV" Interesting? :D There is a lot more to it. I'll continue to show some strange coincidences if anyone is interested. Sorry if I am not explaining this correctly. By now you have probably figured out that I have no background in this. Which is why I am here. Thank you.

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• Binary Trees in Scheme

  - by Javier

  Consider the following BNF defining trees of numbers. Notice that a tree can either be a leaf, a node-1 with one subtrees, or a node-2 with two subtrees. tree ::= (’leaf number) | (’node-1 tree) | (’node-2 tree tree) a. Write a template for recursive procedures on these trees. b. Define the procedure (leaf-count t) that returns the number of leaves in t > (leaf-count ’(leaf 5)) 1 > (leaf-count ’(node-2 (leaf 25) (leaf 17))) 2 > (leaf-count ’(node-1 (node-2 (leaf 4) (node-2 (leaf 2) (leaf 3))))) 3 Here's what I have so far: ;define what a leaf, node-1, and node-2 is (define leaf list) (define node-1 list) (define node-2 list) ;procedure to decide if a list is a leaf or a node (define (leaf? tree) (number? (car tree))) (define (node? tree) (pair? (car tree))) (define (leaf-count tree) (cond ((null? tree) 0) ((number? tree) 0) ((leaf? tree) 1) (else (+ (leaf-count (car tree)) (leaf-count (cdr tree)))))) It looks like it should run just fine, but when I try to run it using a simple test case like (leaf-count '(leaf 5)) I get the following error message: car: expects argument of type pair; given leaf What does this error message mean? I am defining a leaf as a list. But for some reason, it's not seeing that and gives me that error message.

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• Binary Search Tree - Postorder logic

  - by daveb

  I am looking at implementing code to work out binary search tree. Before I do this I was wanting to verify my input data in postorder and preorder. I am having trouble working out what the following numbers would be in postorder and preorder I have the following numbers 4, 3, 14 ,8 ,1, 15, 9, 5, 13, 10, 2, 7, 6, 12, 11, that I am intending to put into an empty binary tree in that order. The order I arrived at for the numbers in POSTORDER is 2, 1, 6, 3, 7, 11, 12, 10, 9, 8, 13, 15, 14, 4. Have I got this right? I was wondering if anyone here would be able to kindly verify if the postorder sequence I came up with is indeed the correct sequence for my input i.e doing left subtree, right subtree and then root. The order I got for pre order (Visit root, do left subtree, do right subtree) is 4, 3, 1, 2, 5, 6, 14 , 8, 7, 9, 10, 12, 11, 15, 13. I can't be certain I got this right. Very grateful for any verification. Many Thanks

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• How to create a Binary Tree from a General Tree?

  - by mno4k

  I have to solve the following constructor for a BinaryTree class in java: BinaryTree(GeneralTree<T> aTree) This method should create a BinaryTree (bt) from a General Tree (gt) as follows: Every Vertex from gt will be represented as a leaf in bt. If gt is a leaf, then bt will be a leaf with the same value as gt If gt is not a leaf, then bt will be constructed as an empty root, a left subTree (lt) and a right subTree (lr). Lt is a stric binary tree created from the oldest subtree of gt (the left-most subtree) and lr is a stric binary tree created from gt without its left-most subtree. The frist part is trivial enough, but the second one is giving me some trouble. I've gotten this far: public BinaryTree(GeneralTree<T> aTree){ if (aTree.isLeaf()){ root= new BinaryNode<T>(aTree.getRootData()); }else{ root= new BinaryNode<T>(null); // empty root LinkedList<GeneralTree<T>> childs = aTree.getChilds(); // Childs of the GT are implemented as a LinkedList of SubTrees child.begin(); //start iteration trough list BinaryTree<T> lt = new BinaryTree<T>(childs.element(0)); // first element = left-most child this.addLeftChild(lt); aTree.DeleteChild(hijos.elemento(0)); BinaryTree<T> lr = new BinaryTree<T>(aTree); this.addRightChild(lr); } } Is this the right way? If not, can you think of a better way to solve this? Thank you!

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• Permuting a binary tree without the use of lists

  - by Banang

  I need to find an algorithm for generating every possible permutation of a binary tree, and need to do so without using lists (this is because the tree itself carries semantics and restraints that cannot be translated into lists). I've found an algorithm that works for trees with the height of three or less, but whenever I get to greater hights, I loose one set of possible permutations per height added. Each node carries information about its original state, so that one node can determine if all possible permutations have been tried for that node. Also, the node carries information on weather or not it's been 'swapped', i.e. if it has seen all possible permutations of it's subtree. The tree is left-centered, meaning that the right node should always (except in some cases that I don't need to cover for this algorithm) be a leaf node, while the left node is always either a leaf or a branch. The algorithm I'm using at the moment can be described sort of like this: if the left child node has been swapped swap my right node with the left child nodes right node set the left child node as 'unswapped' if the current node is back to its original state swap my right node with the lowest left nodes' right node swap the lowest left nodes two childnodes set my left node as 'unswapped' set my left chilnode to use this as it's original state set this node as swapped return null return this; else if the left child has not been swapped if the result of trying to permute left child is null return the permutation of this node else return the permutation of the left child node if this node has a left node and a right node that are both leaves swap them set this node to be 'swapped' The desired behaviour of the algoritm would be something like this: branch / | branch 3 / | branch 2 / | 0 1 branch / | branch 3 / | branch 2 / | 1 0 <-- first swap branch / | branch 3 / | branch 1 <-- second swap / | 2 0 branch / | branch 3 / | branch 1 / | 0 2 <-- third swap branch / | branch 3 / | branch 0 <-- fourth swap / | 1 2 and so on... Sorry for the ridiculisly long and waddly explanation, would really, really apreciate any sort of help you guys could offer me. Thanks a bunch!

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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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• Binary Heap Traversal Proof by Induction

  - by Ahmet Alp Balkan

  How to prove that: Let's say we have two binary heap trees H1 and H2 with size N. If their inorder traversals are the same, prove that H1=H2 by induction on N.

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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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• How to determine if binary tree is balanced?

  - by user69514

  It's been a while from those school years. Got a job as IT specialist at a hospital. Trying to move to do some actual programming now. I'm working on binary trees now, and I was wondering what would be the best way to determine if the tree is height-balanced. I was thinking of something along this: public boolean isBalanced(Node root){ if(root==null){ return true; //tree is empty } else{ int lh = root.left.height(); int rh = root.right.height(); if(lh - rh > 1 || rh - lh > 1){ return false; } } return true; } Is this a good implementation? or am I missing something?

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• java polymorphic binary search tree

  - by ashraf

  how can i implement a polymorphic binary search tree (that makes use of EmptyTree and NonEmptyTree) without using downcasting or class checking?

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