Question

Consider the following BNode class:

class BNode {

int val;

BNode left, right;
...

boolean isBST(BNode u){...}
boolean isBST(int min, BNode u, int max){...}

}

Note that our BNodes do not have parent pointers! It has a method isBST(u) that returns true iff the subtree rooted at u is a BST. Please write the code for isBST(u) and its helper method isBST(min, u, max) which will return false if the keys in the subtree at u do not lie in the range (min, max). The helper method must be recursive.

Answer 1

int isBST(BNode  root)

{

    // if the tree is empty

    if (root == null)

        return true;

    // if the current node disobeys the BST property

    // use the isBST(min, u,max) function to check if each key in left subtree

    // is smaller than root node

    if (root.left != null && this.isBST( Integer.MIN_VALUE , root.left , root.value) )

        return false;

    // if the current node disobeys the BST property

    // use the isBST(min, u,max) function to check if each key in right subtree

    // is greater than root node

    if (root.right != null && this.isBST( root.value , root.right , Integer.MAX_VALUE))

        return false;

    // recursively check if the right and left subtree also are in BST

    if( isBST(root.left) != true || isBST(root.right) != true )

        return false;

    // if the tree is BST

    return true;

}

boolean isBST(int min, BNode u, int max)

{

    if( u == null )

        return true;

   

    // if the current node is outside the range

    if( u.value < min || u.value > max )

        return false;

   

    // recursively check if the right and left subtree also are in BST

    if( isBST(min, root.left, max) != true || isBST(min, root.right, max) != true )

        return false;

   

    return true;

}