Using Java:

Code Skeleton

// node class for a binary tree class BinaryNode { int key; BinaryNode left, right;

// constructor public BinaryNode(int key) { this.key = key; left = right = null; } }

// a binary search tree class public class BinarySearchTree { BinaryNode root;

// constructor public BinarySearchTree() { root = null; }

/* returns the range(difference between the highest and lowest) of values in a binary search tree.*/ int getRange(BinaryNode node) { //write here }

/*Returns the number of nodes in binary search tree that have keys in range [low, high]. range includes low and high*/ int inRangeCount(BinaryNode node, int low, int high) { // write here }

public static void main(String args[]) {

BinarySearchTree bst = new BinarySearchTree(); bst.root = new BinaryNode(3); bst.root.left = new BinaryNode(1); bst.root.right = new BinaryNode(6); bst.root.left.left = new BinaryNode(0); bst.root.left.right = new BinaryNode(2);

System.out.println(bst.getRange(bst.root)); System.out.println(bst.inRangeCount(bst.root,0,3)); } }

Binary Search Tree For this homework, you need to write two methods related to binary search trees. The code skeleton is provided in BinarySearchTree.java file. int getRange (BinaryNode node) returns the range (difference between the highest and lowest) of values in a binary search tree. In the following example (root is the node with key 12), getRange(root) should return 22 -5- 17. You can return -1 if the tree is null. (50 points) 12 10 16 int inRangeCount (BinaryNode node, int low, int high) returns the number of nodes in binary search tree that have keys in range [low, high]. Note that the range includes low and high. In the above example tree, inRangeCount(root, 10,13) should return 3, since there are 3 nodes whose keys are in range [10,13]. (50 points)