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  • Binary Search Tree in Java

    - by John R
    I want to make a generic BST, that can be made up of any data type, but i'm not sure how I could add things to the tree, if my BST is generic. All of my needed code is below. I want my BST made up of Locations, and sorted by the x variable. Any help is appreciated. Major thanks for looking. public void add(E element) { if (root == null) root = element; if (element < root) add(element, root.leftChild); if (element > root) add(element, root.rightChild); else System.out.println("Element Already Exists"); } private void add(E element, E currLoc) { if (currLoc == null) currLoc = element; if (element < root) add(element, currLoc.leftChild); if (element > root) add(element, currLoc.rightChild); else System.out.println("Element Already Exists); } Other Code public class BinaryNode<E> { E BinaryNode; BinaryNode nextBinaryNode; BinaryNode prevBinaryNode; public BinaryNode() { BinaryNode = null; nextBinaryNode = null; prevBinaryNode = null; } } public class Location<AnyType> extends BinaryNode { String name; int x,y; public Location() { name = null; x = 0; y = 0; } public Location(String newName, int xCord, int yCord) { name = newName; x = xCord; y = yCord; } public int equals(Location otherScene) { return name.compareToIgnoreCase(otherScene.name); } }

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  • Java Binary Tree. Priting InOrder traversal

    - by user69514
    I am having some problems printing an inOrder traversal of my binary tree. Even after inserting many items into the tree it's only printing 3 items. public class BinaryTree { private TreeNode root; private int size; public BinaryTree(){ this.size = 0; } public boolean insert(TreeNode node){ if( root == null) root = node; else{ TreeNode parent = null; TreeNode current = root; while( current != null){ if( node.getData().getValue().compareTo(current.getData().getValue()) <0){ parent = current; current = current.getLeft(); } else if( node.getData().getValue().compareTo(current.getData().getValue()) >0){ parent = current; current = current.getRight(); } else return false; if(node.getData().getValue().compareTo(parent.getData().getValue()) < 0) parent.setLeft(node); else parent.setRight(node); } } size++; return true; } /** * */ public void inOrder(){ inOrder(root); } private void inOrder(TreeNode root){ if( root.getLeft() !=null) this.inOrder(root.getLeft()); System.out.println(root.getData().getValue()); if( root.getRight() != null) this.inOrder(root.getRight()); } }

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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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  • 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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  • Effects of changing a node in a binary tree

    - by eSKay
    Suppose I want to change the orange node in the following tree. So, the only other change I'll need to make is in the left pointer of the green node. The blue node will remain the same. Am I wrong somewhere? Because according to this article (that explains zippers), even the blue node needs to be changed. Similarly, in this picture (recolored) from the same article, why do we change the orange nodes at all (when we change x)?

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  • Real world examples of tree structures

    - by Arec Barrwin
    I'm looking for some examples of tree structures that are used in commercial/free software projects, modern or old. I can see examples on wikipedia, but I am looking for more concrete examples and how they're used. For example primary keys in databases are (from what I've read) stored in BST structure or a variation of the BST (feel free to correct me on this) My question isn't limited Binary Search Trees (BSTs), it can include any variation such as red-black, AVL and so on.

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  • Designing binary operations(AND, OR, NOT) in graphs DB's like neo4j

    - by Nicholas
    I'm trying to create a recipe website using a graph database, specifically neo4j using spring-data-neo4j, to try and see what can be done in Graph Databases. My model so far is: (Chef)-[HAS_INGREDIENT]->(Ingredient) (Chef)-[HAS_VALUE]->(Value) (Ingredient)-[HAS_INGREDIENT_VALUE]->(Value) (Recipe)-[REQUIRES_INGREDIENT]->(Ingredient) (Recipe)-[REQUIRES_VALUE]->(Value) I have this set up so I can do things like have the "chef" enter ingredients they have on hand, and suggest recipes, as well as suggest recipes that are close matches, but missing one ingredient. Some recipes can get complex, utilizing AND, OR, and NOT type logic, something like (Milk AND (Butter OR spread OR (vegetable oil OR olive oil))) and I'm wondering if it would be sane to model this in a graph using a tree type representation? An example of what I was thinking is to create three "node" types of AND, OR, and NOT and have each of them connect to the nodes value underneath. How else might this be represented in a Graph Database or is my example above a decent representation?

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  • Understanding binary numbers in terms of real world objects

    - by Kaushik
    When I represent a number in the decimal system, I have an intuitive knowledge of what it amounts to. For example take the number '10': I understand that it means 10 apples or 10 people... i.e I can count in the real world. But as soon as the number is converted to any other system, this understanding no longer applies. For example 10 when converted to binary will be 1010...now what does this represent? Is there a way to understand this number 1010 in terms of counting objects in the real world?

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  • Cannot execute binary file

    - by user291727
    I am new to Ubuntu and I'm trying to install Popcorn Time. I downloaded 32 bit version and tried to install it but that's where the problem started showing. I duble clicked the executable file and well, nothing happened. It's a official download from their web-site but it doesn't work. Maybe I'm doing something wrong.....Anyway, I found out that you can insatll it from a script, but people keep talking in ubuntu terms and I don't understand it, so I have a few questions: 1.How to make a script? (witch I'm suppose to run in terminal using bash comand), 2.Is it normal that i cannot run the installer, and if that is an installer or just files for the program. 3.If it is an installer, how do I make it work? 4.What does " Cannot execute binary file" mean? Thank you in advance, hope I'm not asking too many questions(please understand that I'm new to ubuntu) and sorry about my English. xD

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  • Binary Search Tree Implementation

    - by Gabe
    I've searched the forum, and tried to implement the code in the threads I found. But I've been working on this real simple program since about 10am, and can't solve the seg. faults for the life of me. Any ideas on what I'm doing wrong would be greatly appreciated. BST.h (All the implementation problems should be in here.) #ifndef BST_H_ #define BST_H_ #include <stdexcept> #include <iostream> #include "btnode.h" using namespace std; /* A class to represent a templated binary search tree. */ template <typename T> class BST { private: //pointer to the root node in the tree BTNode<T>* root; public: //default constructor to make an empty tree BST(); /* You have to document these 4 functions */ void insert(T value); bool search(const T& value) const; bool search(BTNode<T>* node, const T& value) const; void printInOrder() const; void remove(const T& value); //function to print out a visual representation //of the tree (not just print the tree's values //on a single line) void print() const; private: //recursive helper function for "print()" void print(BTNode<T>* node,int depth) const; }; /* Default constructor to make an empty tree */ template <typename T> BST<T>::BST() { root = NULL; } template <typename T> void BST<T>::insert(T value) { BTNode<T>* newNode = new BTNode<T>(value); cout << newNode->data; if(root == NULL) { root = newNode; return; } BTNode<T>* current = new BTNode<T>(NULL); current = root; current->data = root->data; while(true) { if(current->left == NULL && current->right == NULL) break; if(current->right != NULL && current->left != NULL) { if(newNode->data > current->data) current = current->right; else if(newNode->data < current->data) current = current->left; } else if(current->right != NULL && current->left == NULL) { if(newNode->data < current->data) break; else if(newNode->data > current->data) current = current->right; } else if(current->right == NULL && current->left != NULL) { if(newNode->data > current->data) break; else if(newNode->data < current->data) current = current->left; } } if(current->data > newNode->data) current->left = newNode; else current->right = newNode; return; } //public helper function template <typename T> bool BST<T>::search(const T& value) const { return(search(root,value)); //start at the root } //recursive function template <typename T> bool BST<T>::search(BTNode<T>* node, const T& value) const { if(node == NULL || node->data == value) return(node != NULL); //found or couldn't find value else if(value < node->data) return search(node->left,value); //search left subtree else return search(node->right,value); //search right subtree } template <typename T> void BST<T>::printInOrder() const { //print out the value's in the tree in order // //You may need to use this function as a helper //and create a second recursive function //(see "print()" for an example) } template <typename T> void BST<T>::remove(const T& value) { if(root == NULL) { cout << "Tree is empty. No removal. "<<endl; return; } if(!search(value)) { cout << "Value is not in the tree. No removal." << endl; return; } BTNode<T>* current; BTNode<T>* parent; current = root; parent->left = NULL; parent->right = NULL; cout << root->left << "LEFT " << root->right << "RIGHT " << endl; cout << root->data << " ROOT" << endl; cout << current->data << "CURRENT BEFORE" << endl; while(current != NULL) { cout << "INTkhkjhbljkhblkjhlk " << endl; if(current->data == value) break; else if(value > current->data) { parent = current; current = current->right; } else { parent = current; current = current->left; } } cout << current->data << "CURRENT AFTER" << endl; // 3 cases : //We're looking at a leaf node if(current->left == NULL && current->right == NULL) // It's a leaf { if(parent->left == current) parent->left = NULL; else parent->right = NULL; delete current; cout << "The value " << value << " was removed." << endl; return; } // Node with single child if((current->left == NULL && current->right != NULL) || (current->left != NULL && current->right == NULL)) { if(current->left == NULL && current->right != NULL) { if(parent->left == current) { parent->left = current->right; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->right; cout << "The value " << value << " was removed." << endl; delete current; } } else // left child present, no right child { if(parent->left == current) { parent->left = current->left; cout << "The value " << value << " was removed." << endl; delete current; } else { parent->right = current->left; cout << "The value " << value << " was removed." << endl; delete current; } } return; } //Node with 2 children - Replace node with smallest value in right subtree. if (current->left != NULL && current->right != NULL) { BTNode<T>* check; check = current->right; if((check->left == NULL) && (check->right == NULL)) { current = check; delete check; current->right = NULL; cout << "The value " << value << " was removed." << endl; } else // right child has children { //if the node's right child has a left child; Move all the way down left to locate smallest element if((current->right)->left != NULL) { BTNode<T>* leftCurrent; BTNode<T>* leftParent; leftParent = current->right; leftCurrent = (current->right)->left; while(leftCurrent->left != NULL) { leftParent = leftCurrent; leftCurrent = leftCurrent->left; } current->data = leftCurrent->data; delete leftCurrent; leftParent->left = NULL; cout << "The value " << value << " was removed." << endl; } else { BTNode<T>* temp; temp = current->right; current->data = temp->data; current->right = temp->right; delete temp; cout << "The value " << value << " was removed." << endl; } } return; } } /* Print out the values in the tree and their relationships visually. Sample output: 22 18 15 10 9 5 3 1 */ template <typename T> void BST<T>::print() const { print(root,0); } template <typename T> void BST<T>::print(BTNode<T>* node,int depth) const { if(node == NULL) { std::cout << std::endl; return; } print(node->right,depth+1); for(int i=0; i < depth; i++) { std::cout << "\t"; } std::cout << node->data << std::endl; print(node->left,depth+1); } #endif main.cpp #include "bst.h" #include <iostream> using namespace std; int main() { BST<int> tree; cout << endl << "LAB #13 - BINARY SEARCH TREE PROGRAM" << endl; cout << "----------------------------------------------------------" << endl; // Insert. cout << endl << "INSERT TESTS" << endl; // No duplicates allowed. tree.insert(0); tree.insert(5); tree.insert(15); tree.insert(25); tree.insert(20); // Search. cout << endl << "SEARCH TESTS" << endl; int x = 0; int y = 1; if(tree.search(x)) cout << "The value " << x << " is on the tree." << endl; else cout << "The value " << x << " is NOT on the tree." << endl; if(tree.search(y)) cout << "The value " << y << " is on the tree." << endl; else cout << "The value " << y << " is NOT on the tree." << endl; // Removal. cout << endl << "REMOVAL TESTS" << endl; tree.remove(0); tree.remove(1); tree.remove(20); // Print. cout << endl << "PRINTED DIAGRAM OF BINARY SEARCH TREE" << endl; cout << "----------------------------------------------------------" << endl; tree.print(); cout << endl << "Program terminated. Goodbye." << endl << endl; } BTNode.h #ifndef BTNODE_H_ #define BTNODE_H_ #include <iostream> /* A class to represent a node in a binary search tree. */ template <typename T> class BTNode { public: //constructor BTNode(T d); //the node's data value T data; //pointer to the node's left child BTNode<T>* left; //pointer to the node's right child BTNode<T>* right; }; /* Simple constructor. Sets the data value of the BTNode to "d" and defaults its left and right child pointers to NULL. */ template <typename T> BTNode<T>::BTNode(T d) : left(NULL), right(NULL) { data = d; } #endif Thanks.

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  • Include Binary Files in DEB package

    - by user22611
    I need to build a DEB package from mainly Node.js Javascript files, but it should include some binary files as well. They are listed inside debian/source/include-binaries. Otherwise I get the error message dpkg-source: error: unrepresentable changes to source The command in question is: bzr builddeb -- -us -uc After adding the file include-binaries, when running bzr builddeb -- -us -uc again, now I get a different error: It says dpkg-source: error: aborting due to unexpected upstream changes, see /tmp/mailadmin_0.0-1.diff.n6m5_6 I have no idea how to get rid of this. In the next line of output it tells me dpkg-source: info: you can integrate the local changes with dpkg-source --commit But if I run this command in the build area of my package, it gives me the unrepresentable changes to source error message again, even though debian/source/include-binaries is present in the build area as well. I am missing the way out of this... I tried deleting all files that are produced by the build process, still no success. Further details: The target directory is /opt/mailadmin. Since this directory is unusual, I listed it in the file debian/mailadmin.install (which contains one line:) opt/mailadmin opt/ The bzr builddeb process uses this file as expected.

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  • In a binary search Tree

    - by user1044800
    In a binary search tree that takes a simple object.....when creating the getter and setter methods for the left, right, and parent. do I a do a null pointer? as in this=this or do I create the object in each method? Code bellow... This is my code: public void setParent(Person parent) { parent = new Person( parent.getName(), parent.getWeight()); //or is the parent supposed to be a null pointer ???? This is the code it came from: public void setParent(Node parent) { this.parent = parent; } Their code takes a node from the node class...my set parent is taking a person object from my person class.....

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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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  • 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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  • Binary Tree in C Insertion Error

    - by Paul
    I'm quite new to C and I'm trying to implement a Binary Tree in C which will store a number and a string and then print them off e.g. 1 : Bread 2 : WashingUpLiquid etc. The code I have so far is: #include <stdio.h> #include <stdlib.h> #define LENGTH 300 struct node { int data; char * definition; struct node *left; struct node *right; }; struct node *node_insert(struct node *p, int value, char * word); void print_preorder(struct node *p); int main(void) { int i = 0; int d = 0; char def[LENGTH]; struct node *root = NULL; for(i = 0; i < 2; i++) { printf("Please enter a number: \n"); scanf("%d", &d); printf("Please enter a definition for this word:\n"); scanf("%s", def); root = node_insert(root, d, def); printf("%s\n", def); } printf("preorder : "); print_preorder(root); printf("\n"); return 0; } struct node *node_insert(struct node *p, int value, char * word) { struct node *tmp_one = NULL; struct node *tmp_two = NULL; if(p == NULL) { p = (struct node *)malloc(sizeof(struct node)); p->data = value; p->definition = word; p->left = p->right = NULL; } else { tmp_one = p; while(tmp_one != NULL) { tmp_two = tmp_one; if(tmp_one->data > value) tmp_one = tmp_one->left; else tmp_one = tmp_one->right; } if(tmp_two->data > value) { tmp_two->left = (struct node *)malloc(sizeof(struct node)); tmp_two = tmp_two->left; tmp_two->data = value; tmp_two->definition = word; tmp_two->left = tmp_two->right = NULL; } else { tmp_two->right = (struct node *)malloc(sizeof(struct node)); tmp_two = tmp_two->right; tmp_two->data = value; tmp_two->definition = word; tmp_two->left = tmp_two->right = NULL; } } return(p); } void print_preorder(struct node *p) { if(p != NULL) { printf("%d : %s\n", p->data, p->definition); print_preorder(p->left); print_preorder(p->right); } } At the moment it seems to work for the ints but the description part only prints out for the last one entered. I assume it has something to do with pointers on the char array but I had no luck getting it to work. Any ideas or advice? 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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  • Need algorithm to add Node in binary tree

    - by m.qayyum
    •if your new element is less or equal than the current node, you go to the left subtree, otherwise to the right subtree and continue traversing •if you arrived at a node, where you can not go any deeper, because there is no subtree, this is the place to insert your new element (5)Root (3)-------^--------(7) (2)---^----(5) ^-----(8) (5)--^ i want to add this last node with data 5...but i can't figure it out...I need a algorithm to do that or in java language

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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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  • 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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  • 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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  • Return the difference between the lowest and highest key

    - by stan
    This is a past exam paper i am attempting and have no way to check if the out put is correct as i am not capable of building one of these things the question is in the title class Tree{ Tree left; Tree right; int key; public static int span(Tree tree) { if ( tree == null ){ return null; } if( tree.left != null) int min = span(tree.left); } if( tree.right != null){ int max = span(tree.right); } return max - min; } } Could anyone suggest what i need to change to get 5/5 marks :D - the only thing we have to do is write the span method, the header was given for us Thanks

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