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

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

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  • binary search tree recursive subtree in java

    - by Art Peterson
    Can anyone point me to a code example (java preferably) or psuedocode that uses recursion to return a subtree that contains all nodes with keys between fromKey and toKey. So if I was to call Tree.subtree(5,10) it should return all nodes in the BST that have keys between 5 and 10 inclusive - but I can't use loops or helper methods...only recursive calls to the subtree method, which takes fromKey and toKey as parameters. Thanks!

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  • Pseudo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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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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  • Building a directory tree from a list of file paths

    - by Abignale
    I am looking for a time efficient method to parse a list of files into a tree. There can be hundreds of millions of file paths. The brute force solution would be to split each path on occurrence of a directory separator, and traverse the tree adding in directory and file entries by doing string comparisons but this would be exceptionally slow. The input data is usually sorted alphabetically, so the list would be something like: C:\Users\Aaron\AppData\Amarok\Afile C:\Users\Aaron\AppData\Amarok\Afile2 C:\Users\Aaron\AppData\Amarok\Afile3 C:\Users\Aaron\AppData\Blender\alibrary.dll C:\Users\Aaron\AppData\Blender\and_so_on.txt From this ordering my natural reaction is to partition the directory listings into groups... somehow... before doing the slow string comparisons. I'm really not sure. I would appreciate any ideas. Edit: It would be better if this tree were lazy loaded from the top down if possible.

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  • Build and render infinite hierarchical category tree from self-referential category table

    - by FreshCode
    I have a Categories table in which each category has a ParentId that can refer to any other category's CategoryId that I want to display as multi-level HTML list, like so: <ul class="tree"> <li>Parent Category <ul> <li>1st Child Category <!-- more sub-categories --> </li> <li>2nd Child Category <!-- more sub-categories --> </li> </ul> </li> </ul> Presently I am recursively rendering a partial view and passing down the next category. It works great, but it's wrong because I'm executing queries in a view. How can I render the list into a tree object and cache it for quick display every time I need a list of all hierarchical categories?

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  • minimax depth first search game tree

    - by Arvind
    Hi I want to build a game tree for nine men's morris game. I want to apply minimax algorithm on the tree for doing node evaluations. Minimax uses DFS to evaluate nodes. So should I build the tree first upto a given depth and then apply minimax or can the process of building the tree and evaluation occur together in recursive minimax DFS? Thank you Arvind

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  • Using an existing IQueryable to create a new dynamic IQueryable

    - by dnoxs
    I have a query as follows: var query = from x in context.Employees where (x.Salary > 0 && x.DeptId == 5) || x.DeptId == 2 order by x.Surname select x; The above is the original query and returns let's say 1000 employee entities. I would now like to use the first query to deconstruct it and recreate a new query that would look like this: var query = from x in context.Employees where ((x.Salary > 0 && x.DeptId == 5) || x.DeptId == 2) && (x,i) i % 10 == 0 order by x.Surname select x.Surname; This query would return 100 surnames. The syntax is probably incorrect, but what I need to do is attach an additional where clause and modify the select to a single field. I've been looking into the ExpressionVisitor but I'm not entirely sure how to create a new query based on an existing query. Any guidance would be appreciated. Thanks you.

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  • Skip List vs. Binary Tree

    - by Claudiu
    I recently came across the data structure known as a Skip list. They seem to have very similar behavior to a binary search tree... my question is - why would you ever want to use a skip list over a binary search tree?

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  • Psuedo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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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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  • What is validating a binary search tree?

    - by dotnetdev
    I read on here of an exercise in interviews known as validating a binary search tree. How exactly does this work? What would one be looking for in validating a binary search tree? I have written a basic search tree, but never heard of this concept. Thanks

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  • An approximate algorithm for finding Steiner Forest.

    - by Tadeusz A. Kadlubowski
    Hello. Consider a weighted graph G=(V,E,w). We are given a family of subsets of vertices V_i. Those sets of vertices are not necessarily disjoint. A Steiner Forest is a forest that for each subset of vertices V_i connects all of the vertices in this subset with a tree. Example: only one subset V_1 = V. In this case a Steiner forest is a spanning tree of the whole graph. Enough theory. Finding such a forest with minimal weight is difficult (NP-complete). Do you know any quicker approximate algorithm to find such a forest with non-optimal weight?

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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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  • 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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  • How to get an hierarchical php structure from a db table, in php array, or JSON

    - by daniel
    Hi guys, can you please help me. How to get an hierarchical php structure from a db table, in php array, or JSON, but with the following format: [{ "attributes" : {"id" : "111"}, "data" : "Some node title", "children" : [ { "attributes" : { "id" : "555"}, "data" : "A sub node title here" } ], "state" : "open" }, { "attributes" : {"id" : "222"}, "data" : "Other main node", "children" : [ { "attributes" : { "id" : "666"}, "data" : "Another sub node" } ], "state" : "open" }] My SQL table contains the fields: ID, PARENT, ORDER, TITLE Can you please help me with this? I'm going crazy trying to get this. Many thanks in advance. Daniel

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  • Binary Search Help

    - by aloh
    Hi, for a project I need to implement a binary search. This binary search allows duplicates. I have to get all the index values that match my target. I've thought about doing it this way if a duplicate is found to be in the middle: Target = G Say there is this following sorted array: B, D, E, F, G, G, G, G, G, G, Q, R S, S, Z I get the mid which is 7. Since there are target matches on both sides, and I need all the target matches, I thought a good way to get all would be to check mid + 1 if it is the same value. If it is, keep moving mid to the right until it isn't. So, it would turn out like this: B, D, E, F, G, G, G, G, G, G (MID), Q, R S, S, Z Then I would count from 0 to mid to count up the target matches and store their indexes into an array and return it. That was how I was thinking of doing it if the mid was a match and the duplicate happened to be in the mid the first time and on both sides of the array. Now, what if it isn't a match the first time? For example: B, D, E, F, G, G, J, K, L, O, Q, R, S, S, Z Then as normal, it would grab the mid, then call binary search from first to mid-1. B, D, E, F, G, G, J Since G is greater than F, call binary search from mid+1 to last. G, G, J. The mid is a match. Since it is a match, search from mid+1 to last through a for loop and count up the number of matches and store the match indexes into an array and return. Is this a good way for the binary search to grab all duplicates? Please let me know if you see problems in my algorithm and hints/suggestions if any. The only problem I see is that if all the matches were my target, I would basically be searching the whole array but then again, if that were the case I still would need to get all the duplicates. Thank you BTW, my instructor said we cannot use Vectors, Hash or anything else. He wants us to stay on the array level and get used to using them and manipulating them.

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  • How to write a recursive function that returns a linked list of nodes, when given a binary tree of n

    - by Jian Lin
    I was once asked of this in an interview: How to write a recursive function that returns a linked list of nodes, when given a binary tree of nodes? (flattening the data) For some reason, I tend to need more than 3 to 5 minutes to solve any recursive problem. Usually, 15 to 20 minutes will be more like it. How could we attack this problem, such as a very systematic way of reaching a solution, so that they can be solved in 3 to 5 minute time frame?

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  • merge sort recursion tree height

    - by Tony
    Hello! I am learning about recursion tree's and trying to figure out how the height of the tree is log b of n where n = 2 and one has 10 elements as input size. I am working with Merge sort. The number of times the split is done is the height of the tree as far as I understood, and the number of levels in the tree is height + 1. But if you take (for merge sort) log2 of 10 you get 1, where if you draw the tree you get at least 2 times that the recursion occurs. Where have I gone wrong? (I hope I am making sense here) NOTE: I am doing a self study, this is not homework!

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  • C++ print out a binary search tree

    - by starcorn
    Hello, Got nothing better to do this Christmas holiday, so I decided to try out making a binary search tree. I'm stuck with the print function. How should the logic behind it work? Since the tree is already inserting it in a somewhat sorted order, and I want to print the tree from smallest values to the biggest. So I need to travel to the furthest left branch of the tree to print the first value. Right, so after that how do I remember the way back up, do I need to save the previous node? A search in wikipedia gave me an solution which they used stack. And other solutions I couldn't quite understand how they've made it, so I'm asking here instead hoping someone can enlight me. I also wonder my insert function is OK. I've seen other's solution being smaller. void treenode::insert(int i) { if(root == 0) { cout << "root" << endl; root = new node(i,root); } else { node* travel = root; node* prev; while(travel) { if(travel->value > i) { cout << "travel left" << endl; prev = travel; travel = travel->left; } else { cout << "travel right" << endl; prev = travel; travel = travel->right; } } //insert if(prev->value > i) { cout << "left" << endl; prev->left = new node(i); } else { cout << "right" << endl; prev->right = new node(i); } } } void treenode::print() { node* travel = root; while(travel) { cout << travel->value << endl; travel = travel->left; } }

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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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