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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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  • 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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  • 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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  • Code to apply expression tree directly to List

    - by Gamma Vega
    Is there a method call in Linq that will apply a expression tree directly on a List<V>? For instance, if I have a expression tree that is built on Order type, and i have a collection of List<Order> items on which i need to apply this expression. I am looking something similar to: class OrderListStore : IQueryable<V>, <other interfaces required to implement custom linq provider> { List<Order> orders; public Expression Expression { get { return Expression.Constant(this); } } IEnumerator<V> IEnumerable<V>.GetEnumerator() { //Here I need to have a method that will take the existing expression //and directly apply on list something like this .. Expression.Translate(orders); } } Any help in this regard is highly appreciated.

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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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  • typesafe NotifyPropertyChanged using linq expressions

    - by bitbonk
    Form Build your own MVVM I have the following code that lets us have typesafe NotifyOfPropertyChange calls: public void NotifyOfPropertyChange<TProperty>(Expression<Func<TProperty>> property) { var lambda = (LambdaExpression)property; MemberExpression memberExpression; if (lambda.Body is UnaryExpression) { var unaryExpression = (UnaryExpression)lambda.Body; memberExpression = (MemberExpression)unaryExpression.Operand; } else memberExpression = (MemberExpression)lambda.Body; NotifyOfPropertyChange(memberExpression.Member.Name); } How does this approach compare to standard simple strings approach performancewise? Sometimes I have properties that change at a very high frequency. Am I safe to use this typesafe aproach? After some first tests it does seem to make a small difference. How much CPU an memory load does this approach potentially induce?

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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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  • How can I generate an Expression tree that queries an object with List<T> as a property?

    - by David Robbins
    Forgive my clumsy explanation, but I have a class that contains a List: public class Document { public int OwnerId { get; set; } public List<User> Users { get; set; } public Document() { } } public class User { public string UserName { get; set; } public string Department { get; set; } } Currently I use PredicateBuilder to perform dynmica queries on my objects. How can I turn the following LINQ statement into an Expression Tree: var predicate= PredicateBuilder.True<User>(); predicate= predicate.And<User>(user => user.Deparment == "HR"); var deptDocs = documents.AsQueryable() .Where(doc => doc.Users .AsQueryable().Count(predicate) > 0) .ToList(); In other words var deptDocs = documents.HasUserAttributes("Department", "HR").ToList();

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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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  • 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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  • Expression Tree with Property Inheritance causes an argument exception

    - by Adam Driscoll
    Following this post: link text I'm trying to create an expression tree that references the property of a property. My code looks like this: public interface IFoo { void X {get;set;} } public interface IBar : IFoo { void Y {get;set;} } public interface IFooBarContainer { IBar Bar {get;set;} } public class Filterer { //Where T = "IFooBarContainer" public IQueryable<T> Filter<T>(IEnumerable<T> collection) { var argument = Expression.Parameter(typeof (T), "item"); //... //where propertyName = "IBar.X"; PropertyOfProperty(argument, propertyName); } private static MemberExpression PropertyOfProperty(Expression expr, string propertyName) { return propertyName.Split('.').Aggregate<string, MemberExpression>(null, (current, property) => Expression.Property(current ?? expr, property)); } } I receive the exception: System.ArgumentException: Instance property 'X' is not defined for type 'IBar' ReSharper turned the code in the link above into the condensed statement in my example. Both forms of the method returned the same error. If I reference IBar.Y the method does not fail.

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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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  • Is a red-black tree my ideal data structure?

    - by Hugo van der Sanden
    I have a collection of items (big rationals) that I'll be processing. In each case, processing will consist of removing the smallest item in the collection, doing some work, and then adding 0-2 new items (which will always be larger than the removed item). The collection will be initialised with one item, and work will continue until it is empty. I'm not sure what size the collection is likely to reach, but I'd expect in the range 1M-100M items. I will not need to locate any item other than the smallest. I'm currently planning to use a red-black tree, possibly tweaked to keep a pointer to the smallest item. However I've never used one before, and I'm unsure whether my pattern of use fits its characteristics well. 1) Is there a danger the pattern of deletion from the left + random insertion will affect performance, eg by requiring a significantly higher number of rotations than random deletion would? Or will delete and insert operations still be O(log n) with this pattern of use? 2) Would some other data structure give me better performance, either because of the deletion pattern or taking advantage of the fact I only ever need to find the smallest item? Update: glad I asked, the binary heap is clearly a better solution for this case, and as promised turned out to be very easy to implement. Hugo

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  • Mutating the expression tree of a predicate to target another type

    - by Jon
    Intro In the application I 'm currently working on, there are two kinds of each business object: the "ActiveRecord" type, and the "DataContract" type. So for example, we have: namespace ActiveRecord { class Widget { public int Id { get; set; } } } namespace DataContracts { class Widget { public int Id { get; set; } } } The database access layer takes care of "translating" between hierarchies: you can tell it to update a DataContracts.Widget, and it will magically create an ActiveRecord.Widget with the same property values and save that. The problem I have surfaced when attempting to refactor this database access layer. The Problem I want to add methods like the following to the database access layer: // Widget is DataContract.Widget interface DbAccessLayer { IEnumerable<Widget> GetMany(Expression<Func<Widget, bool>> predicate); } The above is a simple general-use "get" method with custom predicate. The only point of interest is that I 'm not passing in an anonymous function but rather an expression tree. This is done because inside DbAccessLayer we have to query ActiveRecord.Widget efficiently (LINQ to SQL) and not have the database return all ActiveRecord.Widget instances and then filter the enumerable collection. We need to pass in an expression tree, so we ask for one as the parameter for GetMany. The snag: the parameter we have needs to be magically transformed from an Expression<Func<DataContract.Widget, bool>> to an Expression<Func<ActiveRecord.Widget, bool>>. This is where I haven't managed to pull it off... Attempted Solution What we 'd like to do inside GetMany is: IEnumerable<DataContract.Widget> GetMany( Expression<Func<DataContract.Widget, bool>> predicate) { var lambda = Expression.Lambda<Func<ActiveRecord.Widget, bool>>( predicate.Body, predicate.Parameters); // use lambda to query ActiveRecord.Widget and return some value } This won't work because in a typical scenario, for example if: predicate == w => w.Id == 0; ...the expression tree contains a MemberAccessExpression instance which has a MemberInfo property (named Member) that point to members of DataContract.Widget. There are also ParameterExpression instances both in the expression tree and in its parameter expression collection (predicate.Parameters); After searching a bit, I found System.Linq.Expressions.ExpressionVisitor (its source can be found here in the context of a how-to, very helpful) which is a convenient way to modify an expression tree. Armed with this, I implemented a visitor. This simple visitor only takes care of changing the types in member access and parameter expressions. It may not be complete, but it's fine for the expression w => w.Id == 0. internal class Visitor : ExpressionVisitor { private readonly Func<Type, Type> dataContractToActiveRecordTypeConverter; public Visitor(Func<Type, Type> dataContractToActiveRecordTypeConverter) { this.dataContractToActiveRecordTypeConverter = dataContractToActiveRecordTypeConverter; } protected override Expression VisitMember(MemberExpression node) { var dataContractType = node.Member.ReflectedType; var activeRecordType = this.dataContractToActiveRecordTypeConverter(dataContractType); var converted = Expression.MakeMemberAccess( base.Visit(node.Expression), activeRecordType.GetProperty(node.Member.Name)); return converted; } protected override Expression VisitParameter(ParameterExpression node) { var dataContractType = node.Type; var activeRecordType = this.dataContractToActiveRecordTypeConverter(dataContractType); return Expression.Parameter(activeRecordType, node.Name); } } With this visitor, GetMany becomes: IEnumerable<DataContract.Widget> GetMany( Expression<Func<DataContract.Widget, bool>> predicate) { var visitor = new Visitor(...); var lambda = Expression.Lambda<Func<ActiveRecord.Widget, bool>>( visitor.Visit(predicate.Body), predicate.Parameters.Select(p => visitor.Visit(p)); var widgets = ActiveRecord.Widget.Repository().Where(lambda); // This is just for reference, see below Expression<Func<ActiveRecord.Widget, bool>> referenceLambda = w => w.Id == 0; // Here we 'd convert the widgets to instances of DataContract.Widget and // return them -- this has nothing to do with the question though. } Results The good news is that lambda is constructed just fine. The bad news is that it isn't working; it's blowing up on me when I try to use it (the exception messages are really not helpful at all). I have examined the lambda my code produces and a hardcoded lambda with the same expression; they look exactly the same. I spent hours in the debugger trying to find some difference, but I can't. When predicate is w => w.Id == 0, lambda looks exactly like referenceLambda. But the latter works with e.g. IQueryable<T>.Where, while the former does not (I have tried this in the immediate window of the debugger). I should also mention that when predicate is w => true, it all works just fine. Therefore I am assuming that I 'm not doing enough work in Visitor, but I can't find any more leads to follow on. Can someone point me in the right direction? Thanks in advance for your help!

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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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  • Return nested alias for linq expression

    - by Schotime
    I have the following Linq Expression var tooDeep = shoppers .Where(x => x.Cart.CartSuppliers.First().Name == "Supplier1") .ToList(); I need to turn the name part into the following string. x.Cart.CartSuppliers.Name As part of this I turned the Expression into a string and then split on the . and removed the First() argument. However, when I get to CartSuppliers this returns a Suppliers[] array. Is there a way to get the single type from this. eg. I need to get a Supplier back. Thanks

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  • Returning recursive ternary freaks out

    - by David Titarenco
    Hi, assume this following function: int binaryTree::findHeight(node *n) { if (n == NULL) { return 0; } else { return 1 + max(findHeight(n->left), findHeight(n->right)); } } Pretty standard recursive treeHeight function for a given binary search tree binaryTree. Now, I was helping a friend (he's taking an algorithms course), and I ran into some weird issue with this function that I couldn't 100% explain to him. With max being defined as max(a,b) ((a)>(b)?(a):(b)) (which happens to be the max definition in windef.h), the recursive function freaks out (it runs something like n^n times where n is the tree height). This obviously makes checking the height of a tree with 3000 elements take very, very long. However, if max is defined via templating, like std does it, everything is okay. So using std::max fixed his problem. I just want to know why. Also, why does the countLeaves function work fine, using the same programmatic recursion? int binaryTree::countLeaves(node *n) { if (n == NULL) { return 0; } else if (n->left == NULL && n->right == NULL) { return 1; } else { return countLeaves(n->left) + countLeaves(n->right); } } Is it because in returning the ternary function, the values a => countLeaves(n->left) and b => countLeaves(n->right) were recursively double called simply because they were the resultants? Thank you!

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  • How to create a Binary Tree from a General Tree?

    - by mno4k
    I have to solve the following constructor for a BinaryTree class in java: BinaryTree(GeneralTree<T> aTree) This method should create a BinaryTree (bt) from a General Tree (gt) as follows: Every Vertex from gt will be represented as a leaf in bt. If gt is a leaf, then bt will be a leaf with the same value as gt If gt is not a leaf, then bt will be constructed as an empty root, a left subTree (lt) and a right subTree (lr). Lt is a stric binary tree created from the oldest subtree of gt (the left-most subtree) and lr is a stric binary tree created from gt without its left-most subtree. The frist part is trivial enough, but the second one is giving me some trouble. I've gotten this far: public BinaryTree(GeneralTree<T> aTree){ if (aTree.isLeaf()){ root= new BinaryNode<T>(aTree.getRootData()); }else{ root= new BinaryNode<T>(null); // empty root LinkedList<GeneralTree<T>> childs = aTree.getChilds(); // Childs of the GT are implemented as a LinkedList of SubTrees child.begin(); //start iteration trough list BinaryTree<T> lt = new BinaryTree<T>(childs.element(0)); // first element = left-most child this.addLeftChild(lt); aTree.DeleteChild(hijos.elemento(0)); BinaryTree<T> lr = new BinaryTree<T>(aTree); this.addRightChild(lr); } } Is this the right way? If not, can you think of a better way to solve this? Thank you!

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  • Binary Search Tree - Postorder logic

    - by daveb
    I am looking at implementing code to work out binary search tree. Before I do this I was wanting to verify my input data in postorder and preorder. I am having trouble working out what the following numbers would be in postorder and preorder I have the following numbers 4, 3, 14 ,8 ,1, 15, 9, 5, 13, 10, 2, 7, 6, 12, 11, that I am intending to put into an empty binary tree in that order. The order I arrived at for the numbers in POSTORDER is 2, 1, 6, 3, 7, 11, 12, 10, 9, 8, 13, 15, 14, 4. Have I got this right? I was wondering if anyone here would be able to kindly verify if the postorder sequence I came up with is indeed the correct sequence for my input i.e doing left subtree, right subtree and then root. The order I got for pre order (Visit root, do left subtree, do right subtree) is 4, 3, 1, 2, 5, 6, 14 , 8, 7, 9, 10, 12, 11, 15, 13. I can't be certain I got this right. Very grateful for any verification. Many Thanks

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  • Binary Search Tree node removal

    - by doc
    I've been trying to implement a delete function for a Binary Search Tree but haven't been able to get it to work in all cases. This is my latest attempt: if(t->get_left() == empty) *t = *t->get_left(); else if(t->get_right() == empty) *t = *t->get_right(); else if((t->get_left() != empty) && (t->get_right() != empty)) { Node* node = new Node(t->get_data(), t->get_parent(), t->get_colour(), t->get_left(), t->get_right()); *t = *node; } t is a node and empty is just a node with nothing in it. I'm just trying to swap the values but I'm getting a runtime error. Any ideas? Thanks

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  • New to AVL tree implementation.

    - by nn
    I am writing a sliding window compression algorithm (LZ77) that searches for phrases in a "moving" dictionary. So far I have written a BST where each node is stored in an array and it's index in the array is also the value of the starting position in the window itself. I am now looking at transforming the BST to an AVL tree. I am a little confused at the sample implementations I have seen. Some only appear to store the balance factors whereas others store the height of each tree. Are there any performance advantage/disadvantages of storing the height and/or balance factor for each node? Apologies if this is a very simple question, but I'm still not visualizing how I want to restructure my BST to implement height balancing. Thanks.

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  • To Reference A Generic Method With A Lambda Expression

    - by SDReyes
    It is possible to reference a generic method using a Lambda Expression Object? For example, having: TheObject: public abstract class LambdaExpression : Expression TheMethod (an extension method of LINQ): public static TSource Last<TSource>( this IEnumerable<TSource> source ) I'm trying to create an instance of TheObject, that references to TheMethod. How do you do such thing?

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