Search Results

Search found 4232 results on 170 pages for 'tree traversal'.

Page 2/170 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • 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; } } }

    Read the article

  • 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. 

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • Preoder traversal of a Btree

    - by Phenom
    I'm trying to figure out how to do a preorder traversal of a Btree. I know that generally preorder traversal works like this: preorder(node) { print value in node preorder(left child) preorder(right child) } What's confusing to me is how to make this work with a Btree, since in each node there are multiple values and multiple child pointers. When printing values, do all the values in the node get printed before descending into the left child? Each node looks like this: child1 value1 child2 value2 child3 value3 child4 Also, why would anyone want to do a preorder traversal of a Btree, since an inorder traversal is what will display the values in ascending order?

    Read the article

  • Reasonably faster way to traverse a directory tree in Python?

    - by Sridhar Ratnakumar
    Assuming that the given directory tree is of reasonable size: say an open source project like Twisted or Python, what is the fastest way to traverse and iterate over the absolute path of all files/directories inside that directory? I want to do this from within Python (subprocess is allowed). os.path.walk is slow. So I tried ls -lR and tree -fi. For a project with about 8337 files (including tmp, pyc, test, .svn files): $ time tree -fi > /dev/null real 0m0.170s user 0m0.044s sys 0m0.123s $ time ls -lR > /dev/null real 0m0.292s user 0m0.138s sys 0m0.152s $ time find . > /dev/null real 0m0.074s user 0m0.017s sys 0m0.056s $ tree appears to be faster than ls -lR (though ls -R is faster than tree, but it does not give full paths). find is the fastest. Can anyone think of a faster and/or better approach? On Windows, I may simply ship a 32-bit binary tree.exe or ls.exe if necessary. Update 1: Added find

    Read the article

  • 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; }

    Read the article

  • How do implement a breadth first traversal?

    - by not looking for answer
    //This is what I have. I thought pre-order was the same and mixed it up with depth first! import java.util.LinkedList; import java.util.Queue; public class Exercise25_1 { public static void main(String[] args) { BinaryTree tree = new BinaryTree(new Integer[] {10, 5, 15, 12, 4, 8 }); System.out.print("\nInorder: "); tree.inorder(); System.out.print("\nPreorder: "); tree.preorder(); System.out.print("\nPostorder: "); tree.postorder(); //call the breadth method to test it System.out.print("\nBreadthFirst:"); tree.breadth(); } } class BinaryTree { private TreeNode root; /** Create a default binary tree */ public BinaryTree() { } /** Create a binary tree from an array of objects */ public BinaryTree(Object[] objects) { for (int i = 0; i < objects.length; i++) { insert(objects[i]); } } /** Search element o in this binary tree */ public boolean search(Object o) { return search(o, root); } public boolean search(Object o, TreeNode root) { if (root == null) { return false; } if (root.element.equals(o)) { return true; } else { return search(o, root.left) || search(o, root.right); } } /** Return the number of nodes in this binary tree */ public int size() { return size(root); } public int size(TreeNode root) { if (root == null) { return 0; } else { return 1 + size(root.left) + size(root.right); } } /** Return the depth of this binary tree. Depth is the * number of the nodes in the longest path of the tree */ public int depth() { return depth(root); } public int depth(TreeNode root) { if (root == null) { return 0; } else { return 1 + Math.max(depth(root.left), depth(root.right)); } } /** Insert element o into the binary tree * Return true if the element is inserted successfully */ public boolean insert(Object o) { if (root == null) { root = new TreeNode(o); // Create a new root } else { // Locate the parent node TreeNode parent = null; TreeNode current = root; while (current != null) { if (((Comparable)o).compareTo(current.element) < 0) { parent = current; current = current.left; } else if (((Comparable)o).compareTo(current.element) > 0) { parent = current; current = current.right; } else { return false; // Duplicate node not inserted } } // Create the new node and attach it to the parent node if (((Comparable)o).compareTo(parent.element) < 0) { parent.left = new TreeNode(o); } else { parent.right = new TreeNode(o); } } return true; // Element inserted } public void breadth() { breadth(root); } // Implement this method to produce a breadth first // search traversal public void breadth(TreeNode root){ if (root == null) return; System.out.print(root.element + " "); breadth(root.left); breadth(root.right); } /** Inorder traversal */ public void inorder() { inorder(root); } /** Inorder traversal from a subtree */ private void inorder(TreeNode root) { if (root == null) { return; } inorder(root.left); System.out.print(root.element + " "); inorder(root.right); } /** Postorder traversal */ public void postorder() { postorder(root); } /** Postorder traversal from a subtree */ private void postorder(TreeNode root) { if (root == null) { return; } postorder(root.left); postorder(root.right); System.out.print(root.element + " "); } /** Preorder traversal */ public void preorder() { preorder(root); } /** Preorder traversal from a subtree */ private void preorder(TreeNode root) { if (root == null) { return; } System.out.print(root.element + " "); preorder(root.left); preorder(root.right); } /** Inner class tree node */ private class TreeNode { Object element; TreeNode left; TreeNode right; public TreeNode(Object o) { element = o; } } }

    Read the article

  • In-order tree traversal

    - by Chris S
    I have the following text from an academic course I took a while ago about in-order traversal (they also call it pancaking) of a binary tree (not BST): In-order tree traversal Draw a line around the outside of the tree. Start to the left of the root, and go around the outside of the tree, to end up to the right of the root. Stay as close to the tree as possible, but do not cross the tree. (Think of the tree — its branches and nodes — as a solid barrier.) The order of the nodes is the order in which this line passes underneath them. If you are unsure as to when you go “underneath” a node, remember that a node “to the left” always comes first. Here's the example used (slightly different tree from below) However when I do a search on google, I get a conflicting definition. For example the wikipedia example: Inorder traversal sequence: A, B, C, D, E, F, G, H, I (leftchild,rootnode,right node) But according to (my understanding of) definition #1, this should be A, B, D, C, E, F, G, I, H Can anyone clarify which definition is correct? They might be both describing different traversal methods, but happen to be using the same name. I'm having trouble believing the peer-reviewed academic text is wrong, but can't be certain.

    Read the article

  • 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.

    Read the article

  • Javascript / Jquery Tree Travesal question

    - by Copper
    Suppose I have the following <ul> <li>Item 1</li> <li>Item 2 <ul> <li>Sub Item</li> </ul> </li> <li>Item 3</li> </ul> This list is auto-generated by some other code (so adding exclusive id's/class' is out of the question. Suppose I have some jquery code that states that if I mouseover an li, it gets a background color. However, if I mouseover the "Sub Item" list item, "Item 2" will be highlighted as well. How can I make it so that if the user mouses over "Sub Item" it only puts a background color on that and not on "Item 2" as well?

    Read the article

  • 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; };

    Read the article

  • 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.)

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • reconstructing a tree from its preorder and postorder lists.

    - by NomeN
    Consider the situation where you have two lists of nodes of which all you know is that one is a representation of a preorder traversal of some tree and the other a representation of a postorder traversal of the same tree. I believe it is possible to reconstruct the tree exactly from these two lists, and I think I have an algorithm to do it, but have not proven it. As this will be a part of a masters project I need to be absolutely certain that it is possible and correct (Mathematically proven). However it will not be the focus of the project, so I was wondering if there is a source out there (i.e. paper or book) I could quote for the proof. (Maybe in TAOCP? anybody know the section possibly?) In short, I need a proven algorithm in a quotable resource that reconstructs a tree from its pre and post order traversals. Note: The tree in question will probably not be binary, or balanced, or anything that would make it too easy. Note2: Using only the preorder or the postorder list would be even better, but I do not think it is possible. Note3: A node can have any amount of children. Note4: I only care about the order of siblings. Left or right does not matter when there is only one child.

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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); } }

    Read the article

  • 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; } } }

    Read the article

  • 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 ?

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >