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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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  • Tortoise SVN tree conflict with myself

    - by Jesse Pepper
    Has anyone had the experience of moving a file in tortoise and committing successfully, only to later commit a different change and be told of a tree conflict where: the file in its original location has been deleted, but in tortoise is marked as missing the file in its new location is there, but marked as already added. (I use tortoise SVN, and we have client and server 1.60) Nobody else changed either the directory or the file (according to svn log). Why is this happening? Is there a way to avoid it happening? If it does happen, is there a more elegant way of fixing the problem than by deleting the whole folder and updating again?

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  • B-Tree Revision

    - by stan
    Hi, If we are looking for line intersections (horizontal and vertical lines only) and we have n lines with half of them vertical and no intersections then Sorting the list of line end points on y value will take N log N using mergesort Each insert delete and search of our data structue (assuming its a b-tree) will be < log n so the total search time will be N log N What am i missing here, if the time to sort using mergesort takes a time of N log N and insert and delete takes a time of < log n are we dropping the constant factor to give an overal time of N log N. If not then how comes < log n goes missing in total ONotation run time? Thanks

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  • (External) Java library for creating Tree structure ?

    - by suVasH.....
    I am planning to implement a tree structure where every node has two children and a parent along with various other node properties (and I'd want to do this in Java ) Now, the way to it probably is to create the node such that it links to other nodes ( linked list trick ), but I was wondering if there is any good external library to handle all this low level stuff. ( for eg. the ease of stl::vector vs array in C++ ). I've heard of JDots, but still since i haven't started (and haven't programmed a lot in Java), I'd rather hear out before I begin.

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  • Speech.Recognition GrammarBuilder/Choices Tree Structure

    - by user2210179
    In playing around with C#'s Speech Recognition, I've stumbled across a road block in the creation of an effective GrammerBuilder with Choices (more specifically, Choices of Choices). IE considering the following logical commands. One solution would to "hard code" every combination of Speech lines and add them to a GrammarBuilder (ie "SET LEFT COLOR RED" and "SET RIGHT CLEAR", however, this would quickly max out the limit of 1024, especially when dealing with number combinations. Another solution would to Append all 'columns' as "Choices" (and filter out incorrect paths upon 'recognition', however this seems like it's processor heavy and unnecessary. The middle ground, seems like the best path - with Choices of Choices - like a tree structure on a GrammarBuilder - however I'm not sure how to proceed. Any suggestions?

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  • Unix tree convert to recursive php array

    - by Fordnox
    I have a response from remote server like this: /home/computer/Downloads |-- /home/computer/Downloads/Apple | `-- /home/computer/Downloads/Apple/Pad |-- /home/computer/Downloads/Empty_Folder `-- /home/computer/Downloads/Subfolder |-- /home/computer/Downloads/Subfolder/Empty `-- /home/computer/Downloads/Subfolder/SubSubFolder `-- /home/computer/Downloads/Subfolder/SubSubFolder/Test this is the output for command computer@athome:$ tree -df --noreport -L 5 /home/computer/Downloads/ I would like to parse this string to recursive php array or object, something like this. I would show only part of result to get the idea. array( 'title' => '/home/computer/Downloads', 'children' => array( 0 => array( 'title' => '/home/computer/Downloads/Apple', 'children' => array( ... ) ) ); Response from server can change according to scanned directory. Can someone help me write this function. Please note that this is response from remote server and php functions can not scan any remote dir.

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  • Recursive Binary Search Tree Insert

    - by Nick Sinklier
    So this is my first java program, but I've done c++ for a few years. I wrote what I think should work, but in fact it does not. So I had a stipulation of having to write a method for this call: tree.insertNode(value); where value is an int. I wanted to write it recursively, for obvious reasons, so I had to do a work around: public void insertNode(int key) { Node temp = new Node(key); if(root == null) root = temp; else insertNode(temp); } public void insertNode(Node temp) { if(root == null) root = temp; else if(temp.getKey() <= root.getKey()) insertNode(root.getLeft()); else insertNode(root.getRight()); } Thanks for any advice.

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  • WPF component for 2D tree diagram

    - by pdm2011
    I'm looking for a well-documented, supported WPF component that provides an API for visualisation of 2D tree diagrams. Ideally something easy to use, customisable (i.e. supports various flavours of nodes and splines) and preferably with automated layout control. Tools that look good so far are GoXam (http://www.nwoods.com/components/silverlight-wpf/goxam-overview.htm) and yFiles WPF (http://www.yworks.com/en/products_yfileswpf_about.html). Just wondering if anyone has experience with either of these, or can recommend an alternative? Thanks!

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  • A B+tree simple implementation in C

    - by initpy
    Hi guys, I'm working on a fun project where I need a simple key/value store that uses B+Trees. I studied them some years ago, and to be honest, I don't want to reinvent the wheel, so I'm looking for a simple implementation in C of b+tree that I can just include in my project. I know of sqlite's, dbm's and tokyocabinet's ones but they're a little too "complicated" for my needs. Is there any (even pedagogical) work on this you can refer me to? Do you have some code to share? Thanks a lot!

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  • Flex: change item Style on certain Tree based ItemRenderers

    - by Markus
    Hi Everybody, I have a question concerning Tree items. I want to show where a drop action would be placed... The item will be placed in between two existing elements. So what I want to do is, to take the upper item and draw a line underneath it. But I struggling to address the itemRenderer... I have the index for the itemrenderer, but I dont get a instance of that object. Any help is appreciated! Markus

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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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  • Stuck on solving the Minimal Spanning Tree problem.

    - by kunjaan
    I have reduced my problem to finding the minimal spanning tree in the graph. But I want to have one more constraint which is that the total degree for each vertex shouldnt exceed a certain constant factor. How do I model my problem? Is MST the wrong path? Do you know any algorithms that will help me? One more problem: My graph has duplicate edge weights so is there a way to count the number of unique MSTs? Are there algorithms that do this? Thank You. Edit: By degree, I mean the total number of edges connecting the vertex. By duplicate edge weight I mean that two edges have the same weight.

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  • movedown method not saving new position - cakephp tree

    - by Ryan
    Hi everyone, I am experiencing a problem that has popped up recently and is causing quite a bit of trouble for our system. The app we have relies on using the movedown method to organize content, but as of late it has stopped working and began to generate the following warning: Warning (2): array_values() [<a href='function.array-values'>function.array-values</a>]: The argument should be an array in [/usr/local/home/cake/cake_0_2_9/cake/libs/model/behaviors/tree.php, line 459] The line being referenced: list($node) = array_values($Model->find('first', array( 'conditions' => array($scope, $Model->escapeField() => $id), 'fields' => array($Model->primaryKey, $left, $right, $parent), 'recursive' => $recursive ))); The line calling the method: $this->movedown($id,abs((int)$position)); I have exhausted every idea I could come up with. Has anyone else crossed this issue before? Any help, or pointing in a direction would be much appreciated!

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  • Java : Count even values in a Binary Search Tree recursively

    - by user307682
    Hi, I need to find out how many even values are contained in a binary tree. this is my code. private int countEven(BSTNode root){ if ((root == null)|| (root.value%2==1)) return 0; return 1+ countEven(root.left) + countEven(root.right); } this i just coded as i do not have a way to test this out. I'm not able to test it out at the moment but need an answer so badly. any help is deeply appreciated.

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  • At Last, Red Hat Enterprise Linux 6

    <b>Linux Planet:</b> "Linux vendor Red Hat today released the first public beta of Red Hat Enterprise Linux 6 (RHEL 6), giving observers a look at what's to come in the next version of its flagship operating system platform."

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  • Red Gate Software announces speaker line up for US SQL in the City tour

    SQL in the City is a free, full day training and networking event for database professionals. After the success of last year’s event, Red Gate has expanded the event to cover six cities from sea to shining sea, including: New York, Austin, San Francisco, Chicago, Boston, and Seattle. Compress live data by 73% Red Gate's SQL Storage Compress reduces the size of live SQL Server databases, saving you disk space and storage costs. Learn more.

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  • Rogue black-box java application not responding to standard input redirect

    - by Stefan Kendall
    I have an external java application (blackbox), which requires authentication. I need to run this application in a batch setting, but it seems to be reading from standard input in some nonstandard way. That is, if I set the calling of the program to redirect STDIN to a file (... <password.txt) or pipe data to it (echo mypasword | ...), it does not recognize the input. As I run it, also, it seems to intercept Cntrl+c and Cntrl+d and Cntrl+z as legitimate password characters, so it must be doing something odd and not just reading from standard in. Any idea what this application could be doing to read in input? I need to be able to send it information programmatically, and I'm stumped for the moment.

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  • Black box test cases for insertion procedure

    - by AJ
    insertion_procedure (int a[], int p [], int N) { int i,j,k; for (i=0; i<=N; i++) p[i] = i; for (i=2; i<=N; i++) { k = p[i]; j = 1; while (a[p[j-1]] > a[k]) {p[j] = p[j-1]; j--} p[j] = k; } } What would be few good test cases for this particular insertion procedure?

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  • Generate syntax tree for simple math operations

    - by M28
    I am trying to generate a syntax tree, for a given string with simple math operators (+, -, *, /, and parenthesis). Given the string "1 + 2 * 3": It should return an array like this: ["+", [1, ["*", [2,3] ] ] ] I made a function to transform "1 + 2 * 3" in [1,"+",2,"*",3]. The problem is: I have no idea to give priority to certain operations. My code is: function isNumber(ch){ switch (ch) { case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9': case '.': return true; break; default: return false; break; } } function generateSyntaxTree(text){ if (typeof text != 'string') return []; var code = text.replace(new RegExp("[ \t\r\n\v\f]", "gm"), ""); var codeArray = []; var syntaxTree = []; // Put it in its on scope (function(){ var lastPos = 0; var wasNum = false; for (var i = 0; i < code.length; i++) { var cChar = code[i]; if (isNumber(cChar)) { if (!wasNum) { if (i != 0) { codeArray.push(code.slice(lastPos, i)); } lastPos = i; wasNum = true; } } else { if (wasNum) { var n = Number(code.slice(lastPos, i)); if (isNaN(n)) { throw new Error("Invalid Number"); return []; } else { codeArray.push(n); } wasNum = false; lastPos = i; } } } if (wasNum) { var n = Number(code.slice(lastPos, code.length)); if (isNaN(n)) { throw new Error("Invalid Number"); return []; } else { codeArray.push(n); } } })(); // At this moment, codeArray = [1,"+",2,"*",3] return syntaxTree; } alert('Returned: ' + generateSyntaxTree("1 + 2 * 3"));

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  • How can a B-tree node be represented?

    - by chronodekar
    We're learning B-trees in class and have been asked to implement them in code. The teacher has left choice of programming language to us and I want to try and do it in C#. My problem is that the following structure is illegal in C#, unsafe struct BtreeNode { int key_num; // The number of keys in a node int[] key; // Array of keys bool leaf; // Is it a leaf node or not? BtreeNode*[] c; // Pointers to next nodes } Specifically, one is not allowed to create a pointer to point to the structure itself. Is there some work-around or alternate approach I could use? I'm fairly certain that there MUST be a way to do this within the managed code, but I can't figure it out. EDIT: Eric's answer pointed me in the right direction. Here's what I ended up using, class BtreeNode { public List<BtreeNode> children; // The child nodes public static int MinDeg; // The Minimum Degree of the tree public bool IsLeaf { get; set; } // Is the current node a leaf or not? public List<int> key; // The list of keys ... }

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  • OutOfMemoryError creating a tree recursively?

    - by Alexander Khaos Greenstein
    root = new TreeNode(N); constructTree(N, root); private void constructTree(int N, TreeNode node) { if (N > 0) { node.setLeft(new TreeNode(N-1)); constructTree(N-1, node.getLeft()); } if (N > 1) { node.setMiddle(new TreeNode(N-2)); constructTree(N-2, node.getMiddle()); } if (N > 2) { node.setRight(new TreeNode(N-3)); constructTree(N-3, node.getRight()); } Assume N is the root number, and the three will create a left middle right node of N-1, N-2, N-3. EX: 5 / | \ 4 3 2 /|\ 3 2 1 etc. My GameNode class has the following variables: private int number; private GameNode left, middle, right; Whenever I construct a tree with an integer greater than 28, I get a OutOfMemoryError. Is my recursive method just incredibly inefficient or is this natural? Thanks!

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  • Inside Red Gate - Be Reasonable!

    - by simonc
    As I discussed in my previous posts, divisions and project teams within Red Gate are allowed a lot of autonomy to manage themselves. It's not just the teams though, there's an awful lot of freedom given to individual employees within the company as well. Reasonableness How Red Gate treats it's employees is embodied in the phrase 'You will be reasonable with us, and we will be reasonable with you'. As an employee, you are trusted to do your job to the best of you ability. There's no one looking over your shoulder, no one clocking you in and out each day. Everyone is working at the company because they want to, and one of the core ideas of Red Gate is that the company exists to 'let people do the best work of their lives'. Everything is geared towards that. To help you do your job, office services and the IT department are there. If you need something to help you work better (a third or fourth monitor, footrests, or a new keyboard) then ask people in Information Systems (IS) or Office Services and you will be given it, no questions asked. Everyone has administrator access to their own machines, and you can install whatever you want on it. If there's a particular bit of software you need, then ask IS and they will buy it. As an example, last year I wanted to replace my main hard drive with an SSD; I had a summer job at school working in a computer repair shop, so knew what to do. I went to IS and asked for 'an SSD, a SATA cable, and a screwdriver'. And I got it there and then, even the screwdriver. Awesome. I screwed it in myself, copied all my main drive files across, and I was good to go. Of course, if you're not happy doing that yourself, then IS will sort it all out for you, no problems. If you need something that the company doesn't have (say, a book off Amazon, or you need some specifications printing off & bound), then everyone has a expense limit of £100 that you can use without any sign-off needed from your managers. If you need a company credit card for whatever reason, then you can get it. This freedom extends to working hours and holiday; you're expected to be in the office 11am-3pm each day, but outside those times you can work whenever you want. If you need a half-day holiday on a days notice, or even the same day, then you'll get it, unless there's a good reason you're needed that day. If you need to work from home for a day or so for whatever reason, then you can. If it's reasonable, then it's allowed. Trust issues? A lot of trust, and a lot of leeway, is given to all the people in Red Gate. Everyone is expected to work hard, do their jobs to the best of their ability, and there will be a minimum of bureaucratic obstacles that stop you doing your work. What happens if you abuse this trust? Well, an example is company trip expenses. You're free to expense what you like; food, drink, transport, etc, but if you expenses are not reasonable, then you will never travel with the company again. Simple as that. Everyone knows when they're abusing the system, so simply don't do it. Along with reasonableness, another phrase used is 'Don't be an a**hole'. If you act like an a**hole, and abuse any of the trust placed in you, even if you're the best tester, salesperson, dev, or manager in the company, then you won't be a part of the company any more. From what I know about other companies, employee trust is highly variable between companies, all the way up to CCTV trained on employee's monitors. As a dev, I want to produce well-written & useful code that solves people's problems. Being able to get whatever I need - install whatever tools I need, get time off when I need to, obtain reference books within a day - all let me do my job, and so let Red Gate help other people do their own jobs through the tools we produce. Plus, I don't think I would like working for a company that doesn't allow admin access to your own machine and blocks Facebook! Cross posted from Simple Talk.

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