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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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  • Parsing a given binary tree using python?

    - by kaushik
    Parse a binary tree,referring to given set of features,answering decision tree question at each node to decide left child or right child and find the path to leaf node according to answer given to the decision tree.. input wil be a set of feature which wil help in answering the question at each level to choose the left or right half and the output will be the leaf node.. i need help in implementing this can anyone suggest methods?? Please answer... thanks in advance..

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  • C# TCP socket and binary data

    - by MD
    Hi @All How to send binary data (01110110 for exemple) with C# throught a TCP (using SSL) socket ? I'm using : SslStream.Write() and h[0] = (byte)Convert.ToByte("01110110"); isn't working Thanks.

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  • C syntax or binary optimized syntax?

    - by Dpp
    Let's take an simple example of two lines supposedly doing the same thing: if (value = 96 || value < 0) ... or if (value & ~ 95) ... Say 'If's are costly in a loop of thousands of iterations, is it better to keep with the traditional C syntax or better to find a binary optimized one if possible?

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  • convert a binary file in a list (python)

    - by beratch
    Hi all, I'd like to be able to open a binary file, and make a list (kind of array) with all the chars in, like : "\x21\x23\x22\x21\x22\x31" to ["\x21","\x23","\x22","\x21","\x22","\x31"] What would be the best solution to convert it ? Thanks !

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  • Write binary stream to browser using PHP

    - by Dave Jarvis
    Background Trying to stream a PDF report written using iReport through PHP to the browser. The general problem is: how do you write binary data to the browser using PHP? Working Code The following code does the job, but (for many reasons) it is not as efficient as it should be (the code writes a file then sends the file contents the browser). // Load the MySQL database driver. // java( 'java.lang.Class' )->forName( 'com.mysql.jdbc.Driver' ); // Attempt a database connection. // $conn = java( 'java.sql.DriverManager' )->getConnection( "jdbc:mysql://localhost:3306/climate?user=$user&password=$password" ); // Extract parameters. // $params = new java('java.util.HashMap'); $params->put('DistrictCode', '101'); $params->put('StationCode', '0066'); $params->put('CategoryCode', '010'); // Use the fill manager to produce the report. // $fm = java('net.sf.jasperreports.engine.JasperFillManager'); $pm = $fm->fillReport($report, $params, $conn); header('Cache-Control: no-cache private'); header('Content-Description: File Transfer'); header('Content-Disposition: attachment, filename=climate-report.pdf'); header('Content-Type: application/pdf'); header('Content-Transfer-Encoding: binary'); header('Content-Length: ' . strlen( $result ) ); $path = realpath( "." ) . "/output.pdf"; $em = java('net.sf.jasperreports.engine.JasperExportManager'); $result = $em->exportReportToPdfFile($pm,$path); readfile( $path ); $conn->close(); Non-working Code To remove the slight redundancy (i.e., write directly to the browser), the following code looks like it should work, but it does not: $em = java('net.sf.jasperreports.engine.JasperExportManager'); $result = $em->exportReportToPdf($pm); header('Content-Length: ' . strlen( $result ) ); echo $result; Content is sent to the browser, but the file is corrupt (it begins with the PDF header) and cannot be read by any PDF reader. Question How can I take out the middle step of writing to the file and write directly to the browser so that the PDF is not corrupted? Thank you!

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  • reading binary datafile and writing into decimal no file

    - by swaroop b banerjee
    exp data is generated by my mc scaler card as a binary file with first 511 bytes as header and then 24 bit data followed by four bit roi data. i am not a expert in programming. i do understand a little. I would like to convert this file into a file (without header) decimal nos with first col as channel no (1 to 8191) then the data (24 bit) then the Roi data (4 bit). I am looking for source code in c or qbasic. thanks

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  • Free-as-in-beer binary file format inspector

    - by fbrereto
    I am looking for a utility that gives me the ability to specify a binary file format and then interpret a file of bytes according to that format. (Something along the lines of the 010 Editor, but infinitely more cost-effective). Something that runs on Mac OS X would be preferred, but I'm interested to see what all is out there in general (while more of a hassle I'd be willing to run a tool on Windows if it were superior.) What's your preference?

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  • How do you display a binary search tree?

    - by fakeit
    I'm being asked to display a binary search tree in sorted order. The nodes of the tree contain strings. I'm not exactly sure what the best way is to attack this problem. Should I be traversing the tree and displaying as I go? Should I flatten the tree into an array and then use a sorting algorithm before I display? I'm not looking for the actual code, just a guide where to go next.

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  • Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

    - by Justin L.
    It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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  • Binary to Date (C#) 64 Bit Format

    - by Veskechky
    We have a binary file from which we have identified the following dates (as Int64). We now the following facts about the Date/Time format; The 64 bit Date has a resolution to the microsecond The 64 bit Date has a range of 4095 years The Int64 9053167636875050944 (0x7DA34FFFFFFFFFC0) = 9th March 2010 The Int64 9053176432968073152 (0x7DA357FFFFFFFFC0) = 10th March 2010 The Int64 9053185229061095360 (0x7DA35FFFFFFFFFC0) = 11th March 2010 The Int64 9053194025154117568 (0x7DA367FFFFFFFFC0) = 12th March 2010 Any help on figuring out a way to convert this to a valid C# Date/Time is appreciated.

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  • Convert 64bit Binary to Long equivalent

    - by washtik
    How can we convert the following 64 bit binary into the long equivalent; 01111101 10100011 01001111 11111111 11111111 11111111 11111111 11000000 equals 7D A3 4F FF FF FF FF C0 HEX equals 9053167636875050944 << this is the value we want in a C# variable

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  • C++ exceptions binary compatibility

    - by aaa
    hi. my project uses 2 different C++ compilers, g++ and nvcc (cuda compiler). I have noticed exception thrown from nvcc object files are not caught in g++ object files. are C++ exceptions supposed to be binary compatible in the same machine? what can cause such behavior?

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  • Binary file reading problem

    - by ScReYm0
    Ok i have problem with my code for reading binary file... First i will show you my writing code: void book_saving(char *file_name, struct BOOK *current) { FILE *out; BOOK buf; out = fopen(file_name, "wb"); if(out != NULL) { printf_s("Writting to file..."); do { if(current != NULL) { strcpy(buf.catalog_number, current->catalog_number); strcpy(buf.author, current->author); buf.price = current->price; strcpy(buf.publisher, current->publisher); strcpy(buf.title, current->title); buf.price = current->year_published; fwrite(&buf, sizeof(BOOK), 1, out); } current = current->next; }while(current != NULL); printf_s("Done!\n"); fclose(out); } } and here is my "version" for reading it back: int book_open(struct BOOK *current, char *file_name) { FILE *in; BOOK buf; BOOK *vnext; int count; int i; in = fopen("west", "rb"); printf_s("Reading database from %s...", file_name); if(!in) { printf_s("\nERROR!"); return 1; } i = fread(&buf,sizeof(BOOK), 1, in); while(!feof(in)) { if(current != NULL) { current = malloc(sizeof(BOOK)); current->next = NULL; } strcpy(current->catalog_number, buf.catalog_number); strcpy(current->title, buf.title); strcpy(current->publisher, buf.publisher); current->price = buf.price; current->year_published = buf.year_published; fread(&buf, 1, sizeof(BOOK), in); while(current->next != NULL) current = current->next; fclose(in); } printf_s("Done!"); return 0; } I just need to save my linked list in binary file and to be able to read it back ... please help me. The program just don't read it or its crash every time different situation ...

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  • binary files writing/reading problems...

    - by ScReYm0
    Ok i have problem with my code for reading binary file... First i will show you my writing code: void book_saving(char *file_name, struct BOOK *current) { FILE *out; BOOK buf; out = fopen(file_name, "wb"); if(out != NULL) { printf_s("Writting to file..."); do { if(current != NULL) { strcpy(buf.catalog_number, current-catalog_number); strcpy(buf.author, current-author); buf.price = current-price; strcpy(buf.publisher, current-publisher); strcpy(buf.title, current-title); buf.price = current-year_published; fwrite(&buf, sizeof(BOOK), 1, out); } current = current-next; }while(current != NULL); printf_s("Done!\n"); fclose(out); } } and here is my "version" for reading: int book_open(struct BOOK *current, char *file_name) { FILE *in; BOOK buf; BOOK *vnext; int count; int i; in = fopen("west", "rb"); printf_s("Reading database from %s...", file_name); if(!in) { printf_s("\nERROR!"); return 1; } i = fread(&buf,sizeof(BOOK), 1, in); while(!feof(in)) { if(current != NULL) { current = malloc(sizeof(BOOK)); current-next = NULL; } strcpy(current-catalog_number, buf.catalog_number); strcpy(current-title, buf.title); strcpy(current-publisher, buf.publisher); current-price = buf.price; current-year_published = buf.year_published; fread(&buf, 1, sizeof(BOOK), in); while(current-next != NULL) current = current-next; fclose(in); } printf_s("Done!"); return 0; } I just need to save my linked list in binary file and to be able to read it back ... please help me. The program just don't read it or its crash every time different situation ...

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  • Maintaing list order in binary tree.

    - by TheBigO
    Given a sequence of numbers, I want to insert the numbers into a balanced binary tree such that when I do a preorder traversal on the tree, it gives me the sequence back. How can I construct the insert method corresponding to this requirement? Remember that the tree must be balanced, so there isn't a completely trivial solution. I was trying to do this with a modified version of an AVL tree, but I'm not sure if this can work out.

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  • Balancing a Binary Tree (AVL)

    - by Gustavo Carreno
    Ok, this is another one in the theory realm for the CS guys around. In the 90's I did fairly well in implementing BST's. The only thing I could bever get my head around was the intricacy of the algorithm to balance a Binary Tree (AVL). Can you guys help me on this?

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