Hackerrank: Is This a Binary Tree?

Question

Problem Statement: https://www.hackerrank.com/challenges/is-binary-search-tree

My Solution:

boolean checkBST(Node root) {
    if(root == null || (root.left == null && root.right == null)) {
        return true;
    } else if(root.left != null && root.right == null) {
        return (root.data > root.left.data) && checkBST(root.left);
    } else if(root.right != null && root.left == null) {
        return (root.data < root.right.data) && checkBST(root.right);
    } else {
        return (root.data > root.left.data) && (root.data < root.right.data) && checkBST(root.left) && checkBST(root.right);
    }
}

Getting "Wrong Answer" for few of the test cases. I know there are a lot of ways to solve this problem, but I am trying to figure out the error in the above solution.

Note: I am not able to debug for those specific test cases because the code doesn't show how the BST is built with those testcases.

Edit: working solution:

  boolean bstUtil(Node root, int min, int max) {
            return root == null 
                || (root.data > min && root.data < max) 
                && bstUtil(root.left, min, root.data) 
                && bstUtil(root.right, root.data, max);
   }

   boolean checkBST(Node root) {
        return bstUtil(root, -1, 10001);
   }

Answer 1

Consider the following tree :

             3
            / \
           2   5
          / \
         1   6

Your test would return true, even though this is not a binary search tree (since the left sub-tree of the root contains a node (6) larger than the root (3)).

It's not enough to check that the left sub-tree is a binary search tree and the left child is smaller than the root. You must also check that every node of the left child is smaller than the root.