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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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• 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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• 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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• 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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• 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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• Is there an existing solution to the multithreaded data structure problem?

  - by thr

  I've had the need for a multi-threaded data structure that supports these claims: Allows multiple concurrent readers and writers Is sorted Is easy to reason about Fulfilling multiple readers and one writer is a lot easier, but I really would wan't to allow multiple writers. I've been doing research into this area, and I'm aware of ConcurrentSkipList (by Lea based on work by Fraser and Harris) as it's implemented in Java SE 6. I've also implemented my own version of a concurrent Skip List based on A Provably Correct Scalable Concurrent Skip List by Herlihy, Lev, Luchangco and Shavit. These two implementations are developed by people that are light years smarter then me, but I still (somewhat ashamed, because it is amazing work) have to ask the question if these are the two only viable implementations of a concurrent multi reader/writer data structures available today?

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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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• check if a tree is a binary search tree

  - by TimeToCodeTheRoad

  I have written the following code to check if a tree is a Binary search tree. Please help me check the code: Pair p{ boolean isTrue; int min; int max; } public boo lean isBst(BNode v){ return isBST1(v).isTrue; } public Pair isBST1(BNode v){ if(v==null) return new Pair(true, INTEGER.MIN,INTEGER.MAX); if(v.left==null && v.right==null) return new Pair(true, v.data, v.data); Pair pLeft=isBST1(v.left); Pair pRight=isBST1(v.right); boolean check=pLeft.max<v.data && v.data<= pRight.min; Pair p=new Pair(); p.isTrue=check&&pLeft.isTrue&&pRight.isTrue; p.min=pLeft.min; p.max=pRight.max; return p; } Note: This function checks if a tree is a binary search tree

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• Constructing a Binary Tree from its traversals

  - by user991710

  I'm trying to construct a binary tree (unbalanced), given its traversals. I'm currently doing preorder + inorder but when I figure this out postorder will be no issue at all. I realize there are some question on the topic already but none of them seemed to answer my question. I've got a recursive method that takes the Preorder and the Inorder of a binary tree to reconstruct it, but is for some reason failing to link the root node with the subsequent children. Note: I don't want a solution. I've been trying to figure this out for a few hours now and even jotted down the recursion on paper and everything seems fine... so I must be missing something subtle. Here's the code: public static <T> BinaryNode<T> prePlusIn( T[] pre, T[] in) { if(pre.length != in.length) throw new IllegalArgumentException(); BinaryNode<T> base = new BinaryNode(); base.element = pre[0]; // * Get root from the preorder traversal. int indexOfRoot = 0; if(pre.length == 0 && in.length == 0) return null; if(pre.length == 1 && in.length == 1 && pre[0].equals(in[0])) return base; // * If both arrays are of size 1, element is a leaf. for(int i = 0; i < in.length -1; i++){ if(in[i].equals(base.element)){ // * Get the index of the root indexOfRoot = i; // in the inorder traversal. break; } // * If we cannot, the tree cannot be constructed as the traversals differ. else throw new IllegalArgumentException(); } // * Now, we recursively set the left and right subtrees of // the above "base" root node to whatever the new preorder // and inorder traversals end up constructing. T[] preleft = Arrays.copyOfRange(pre, 1, indexOfRoot + 1); T[] preright = Arrays.copyOfRange(pre, indexOfRoot + 1, pre.length); T[] inleft = Arrays.copyOfRange(in, 0, indexOfRoot); T[] inright = Arrays.copyOfRange(in, indexOfRoot + 1, in.length); base.left = prePlusIn( preleft, inleft); // * Construct left subtree. base.right = prePlusIn( preright, inright); // * Construc right subtree. return base; // * Return fully constructed tree } Basically, I construct additional arrays that house the pre- and inorder traversals of the left and right subtree (this seems terribly inefficient but I could not think of a better way with no helpers methods). Any ideas would be quite appreciated. Side note: While debugging it seems that the root note never receives the connections to the additional nodes (they remain null). From what I can see though, that should not happen... EDIT: To clarify, the method is throwing the IllegalArgumentException @ line 21 (else branch of the for loop, which should only be thrown if the traversals contain different elements.

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• Log 2 N generic comparison tree

  - by Morano88

  Hey! I'm working on an algorithm for Redundant Binary Representation (RBR) where every two bits represent a digit. I designed the comparator that takes 4 bits and gives out 2 bits. I want to make the comparison in log 2 n so If I have X and Y .. I compare every 2 bits of X with every 2 bits of Y. This is smooth if the number of bits of X or Y equals n where (n = 2^X) i.e n = 2,4,8,16,32,... etc. Like this : However, If my input let us say is 6 or 10 .. then it becomes not smooth and I have to take into account some odd situations like this : I have a shallow experience in algorithms .. is there a generic way to do it .. so at the end I get only 2 bits no matter what the input is ? I just need hints or pseudo-code. If my question is not appropriate here .. so feel free to flag it or tell me to remove it. I'm using VHDL by the way!

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• Finding height in Binary Search Tree

  - by mike

  Hey I was wondering if anybody could help me rework this method to find the height of a binary search tree. So far my code looks like this however the answer im getting is larger than the actual height by 1, but when I remove the +1 from my return statements its less than the actual height by 1? I'm still trying to wrap my head around recursion with these BST any help would be much appreciated. public int findHeight(){ if(this.isEmpty()){ return 0; } else{ TreeNode<T> node = root; return findHeight(node); } } private int findHeight(TreeNode<T> aNode){ int heightLeft = 0; int heightRight = 0; if(aNode.left!=null) heightLeft = findHeight(aNode.left); if(aNode.right!=null) heightRight = findHeight(aNode.right); if(heightLeft > heightRight){ return heightLeft+1; } else{ return heightRight+1; } }

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• nHibernate: Query tree nodes where self or ancestor matches condition

  - by Famous Nerd

  I have see a lot of competing theories about hierarchical queries in fluent-nHibernate or even basic nHibernate and how they're a difficult beast. Does anyone have any knowledge of good resources on the subject. I find myself needing to do queries similar to: (using a file system analog) select folderObjects from folders where folder.Permissions includes :myPermissionLevel or [any of my ancestors] includes :myPermissionLevel This is a one to many tree, no node has multiple parents. I'm not sure how to describe this in nHibernate specific terms or, even sql-terms. I've seen the phrase "nested sets" mentioned, is this applicable? I'm not sure. Can anyone offer any advice on approaches to writing this sort of nHibernate query?

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• An Efficient data structure for Sorted List

  - by holydiver

  I want to save my objects according to a key in the attributes of my object in a sorted fashion. Later on I'll access these objects sequentially from max key to min key. I'll do some search tasks as well. I consider to use either AVL tree or RB Tree. As far as i know they are nearly equivalent in theory(Both have O(logn)). But in practice which one might be better in performance in my situation. And is there a better alternative than those, considering that I'll mostly do insert and sequentially access to the ds.

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• Using recursion to to trim a binary tree based on a given min and max value

  - by Justin

  As the title says, I have to trim a binary tree based on a given min and max value. Each node stores a value, and a left/right node. I may define private helper methods to solve this problem, but otherwise I may not call any other methods of the class nor create any data structures such as arrays, lists, etc. An example would look like this: overallRoot _____[50]____________________ / \ __________[38] _______________[90] / \ / _[14] [42] [54]_____ / \ \ [8] [20] [72] \ / \ [26] [61] [83] trim(52, 65); should return: overallRoot [54] \ [61] My attempted solution has three methods: public void trim(int min, int max) { rootFinder(overallRoot, min, max); } First recursive method finds the new root perfectly. private void rootFinder(IntTreeNode node, int min, int max) { if (node == null) return; if (overallRoot.data < min) { node = overallRoot = node.right; rootFinder(node, min, max); } else if (overallRoot.data > max) { node = overallRoot = node.left; rootFinder(node, min, max); } else cutter(overallRoot, min, max); } This second method should eliminate any further nodes not within the min/max, but it doesn't work as I would hope. private void cutter(IntTreeNode node, int min, int max) { if (node == null) return; if (node.data <= min) { node.left = null; } if (node.data >= max) { node.right = null; } if (node.data < min) { node = node.right; } if (node.data > max) { node = node.left; } cutter(node.left, min, max); cutter(node.right, min, max); } This returns: overallRoot [54]_____ \ [72] / [61] Any help is appreciated. Feel free to ask for further explanation as needed.

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• Help With Lisp Code for a Binary Tree

  - by iulia

  I have (setq l2 '(1 (2 b (c 1 b))(a (1 2) d))) ( defun drumuri (l3) ( cond ( (atom l3) ( cons l3 nil)) ( t ( append ( cons ( car l3 ) nil) ( drumuri ( cadr l3)) (cons (car l3)nil) ( drumuri ( caddr l3)) )))) ( drumuri l2) and it gives me: Break 2 [4]> DRUMURI Break 2 [4]> (1 2 B 2 C 1 C B 1 A 1 2 1 NIL A D) but i need: ((1 2 B)(1 2 C 1)(1 2 C B)(1 A 1 2)(1 A D))

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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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• Java Binary Tree. Priting InOrder traversal

  - by user69514

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

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• 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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• CompareTo may return 0, alternative to TreeSet/TreeMap

  - by Bart van Heukelom

  I need a sorted set of objects and am currently using the TreeSet. My problem is that the compareTo of the objects will often return 0, meaning the order of those two objects is to be left unchanged. TreeMap (used by TreeSet by default) will then regard them as the same object, which is not true. What alternative to TreeMap can I use?

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• How to create a complete binary tree of height 'h' using Python?

  - by Jack

  Here is the node structure class Node: def __init__(self, data): # initializes the data members self.left = None self.right = None self.parent = None self.data = data complete binary tree Definition: A binary tree in which every level, except possibly the deepest, is completely filled. At depth n, the height of the tree, all nodes must be as far left as possible. -- http://www.itl.nist.gov/div897/sqg/dads/HTML/completeBinaryTree.html I am looking for an efficient algorithm.

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• How to print a binary tree of functions/terminals like a lisp statement?

  - by sieve411

  I have a binary tree of functions and terminal values. I'd like to print this tree as a lisp statement would be represented! For example, a tree with just a root of "+" and terminals of "2" and 4" would read (+ (2 4)).

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• Finding the largest subtree in a BST

  - by rakeshr

  Given a binary tree, I want to find out the largest subtree which is a BST in it. Naive approach: I have a naive approach in mind where I visit every node of the tree and pass this node to a isBST function. I will also keep track of the number of nodes in a sub-tree if it is a BST. Is there a better approach than this ?

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• How to create Binary Tree in a image(.jpeg)

  - by abc

  I have data structured in Binary tree format, i want to represent it into an image(*. jpeg) then i need to display that image on web page and all the data will come @ runtime, so image processing should be done @ runtime, how to do this ? This is what my thought solution any other suitable solution are also welcomed, web site is in .NET , i am thinking to produce image using java api then integrate it to .NET wither through WEB-SERVICE call or any other solutions are also welcomed.

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• g_tree_insert overwrites all data

  - by pechenie

  I wonder how I should use the GTree (from GLib) to store data? Every new value I insert into GTree with g_tree_insert routine is overwrite the previous one! GTree *tree; //init tree = g_tree_new( g_str_equal ); //"g_str_equal" is a GLib default compare func //... for( i = 0; i < 100; ++i ) g_tree_insert( tree, random_key(), random_value() ); //insert some random vals // printf( "%d", g_tree_nnodes( tree ) ); //should be 100? NO! Prints "1"!!! What am I doing wrong? Thank you.

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• Java equivalent of C++ std::map?

  - by Rudiger

  I'm looking for a Java class with the characteristics of C++ std::map's usual implementation (as I understand it, a self-balancing binary search tree): O(log n) performance for insertion/removal/search Each element is composed of a unique key and a mapped value Keys follow a strict weak ordering I'm looking for implementations with open source or design documents; I'll probably end up rolling my own support for primitive keys/values. This question's style is similar to: Java equivalent of std::deque, whose answer was "ArrayDeque from Primitive Collections for Java".

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