binary-search-tree

I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs. //Assume that nodes are represented by numbers from 1...n where 1=root and even //nos.=left child and odd nos=right child. public int parent(Node node){ if (node % 2 == 0){ if (root.left==node) return root; else return parent(root.left); } //same case for right } But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right . If it's the child or if the node that was queried else ,

I have built a splay tree and I am trying to print it out reverse in order so that when you turn your head to the left you can see the tree in a normal manner. I have written the following code and it outputs the tree somewhat correctly but it adds extra spaces in on the rightmost node and it doesn't add spaces for all of the child nodes that should be placed below the root node: public void printReverseInOrder() { if (root != null) { reverseInOrder(root, 0); } else { System.out.println(); } } public void reverseInOrder(BSTnode h, int indent) { if (h != null) { for (int i = 0; i < indent; i++)

My problem is as follows: I have a binary search tree with keys: a1<a2<...<an , the problem is to print all the (a_i, a_i+1) pairs in the tree (where i={1,2,3,...}) using a recursive algorithm in O(n) time without any global variable and using O(1) extra space. An example: Let the keys be: 1,2, ..., 5 Pairs that will be printed: (1,2) (2,3) (3, 4) (4, 5) So you can't do inorder traversal in the tree and find the successor/predecessor for each node. Because that would take O(nh) time and if the tree is balanced, it will be O(nlgn) for the whole tree. Although you are right that finding an in

I am writing a program to traverse a binary search tree.Here's my code: Main.java public class Main { public static void main(String[] args) { BinaryTree binaryTree = new BinaryTree(); binaryTree.add(50); binaryTree.add(40); binaryTree.add(39); binaryTree.add(42); binaryTree.add(41); binaryTree.add(43); binaryTree.add(55); binaryTree.add(65); binaryTree.add(60); binaryTree.inOrderTraversal(binaryTree.root); } } Node.java public class Node { int data; Node left; Node right; Node parent; public Node(int d) { data = d; left = null; right = null; } } BinaryTree.java public class BinaryTree { Node

I have a trouble understanding what's a successor of a node X whenever it doesn't have a right child. From what I had understood, if a node X had no right child, then it would not have a successor. But my textbook says the following: If the right sub-tree of node X is empty and X has a successor Y ... How can X have a successor when it has not right child? The successor is simply the next element in the ordered sequence; it doesn't necessarily have to be a child element. For example, the successor of 5 below is 7 : 7 / \ 5 8 Manali The successor of a node X is the smallest greater element S in