binary-tree

I am writing an Add function to add nodes to a binary tree non recursively. I have run into the problem of only being able to produce one level deep binary tree. I debugged it and I know where the problem is but can't figure out how to fix it. Maybe fresh pair of eyes will see something i dont... The problem is that my temp node is being reset to the root value every new function call and hence, adding nodes linearly. Anyways, here is the function: void BinaryTreeNonRec::add(int num){ treeNode *temp, *parent; if (root==NULL) { root = new treeNode; root->data=num; root->left=0; root->right=0;

I have just started studying Binary Tree. Is there an algorithm to find out the binary tree structure,given the Inorder and Postorder OR Inorder and Preorder? I have been trying to do it manually,but it never comes out correct.For eg.-These two are valid Inorder and Postorder traversal of a given tree: Inorder: D B F E A G C L J H K Postorder : D F E B G L J K H C A Clearly A is the root as it is the last element in Postorder. Now looking in Inorder,the left subtree becomes: {D B F E} and right subtree becomes: {G C L J H K}. The root of right subtree would be the second last element in

I am trying to create a binary tree for use in a knockout tournament. The tree consists of TNodes with Left and Right pointers. This is the code that I have come up with (below); however, it runs into difficulties with the pointers in the CreateTree section. Once this creates an empty tree of large enough size, I need to add the names on the Memo1.List to the bottoms of the tree so I can pair them up for matches. How would I do this? Type TNodePtr = ^TNode; TNode = Record Data:String; Left:TNodePtr; Right:TNodePtr; end; Type TTree = Class Private Root:TNodePtr; Public Function GetRoot:TNodePtr

I was trying to understand how it is so intuitive to implement postorder traversal using 2 stacks. How did someone come up with it, is it just an observation or some particular way of thinking which helps one come up with such methods. If yes then please explain how to think in the right direction. Let me explain how I stumbled on the solution: You start off with this simple observation: prima facie, the pre-order and post-order traversal differ only in their order of operations : preOrder(node): if(node == null){ return } visit(node) preOrder(node.leftChild) preOrder(node.rightChild)

I've been working on an assignment and now I'm stuck with buggy destructors. I have to create a generic binary tree with all the usual member functions and some special operators. There's also a restriction: everything must work iteratively so no nasty recursive hacks this time. There is obviously something very wrong with the destructor of BinTreeNode class because if I delete the node like this: BinTreeNode<int> * node = new BinTreeNode<int>(); delete node; I can still access its data: node->getData(); //should fail miserably so deletion has no effect but I have no usable idea how I should

Need help finding a way to construct a binary tree given the inorder and level traversal. Is it possible to do it using recursion since the level traversal has to be done by using a queue? Here's how you can approach this problem. It's easier to think of how to approach each step by looking from reverse: 8 / \ 4 9 / \ \ 2 6 10 / 1 You have the following: Inorder: 1 2 4 6 8 9 10 Level: 8 4 9 2 6 10 1 Level 1 - Root A level traversal is a left to right, top to down traversal of the tree (like breadth-first search). In this example, you know that 8 will be the root node. Now looking at the

The rules for a Red-Black Tree: Every node is either red or black. The root is black. Every leaf (NIL) is black. If a node is red, then both its children are black. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes. Rule 4 mentions that red nodes need both black childs but what if there is just one child to begin with? Is there an argument to prove or disprove this? No,a red node cannot have one child,consider the following cases:- 1.If the single child it has is red...this cannot happen because no two consecutive nodes can be red. 2.If