Check for BST [MEDIUM]

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Question: 

Given a binary tree, return true if it is BST, else false. For example, the following tree is not BST, because 11 is in left subtree of 10.

        10
     /     \
   7       39
     \
      11

Input:
The task is to complete the method which takes one argument, root of Binary Tree. The struct Node has a data part which stores the data, pointer to left child and pointer to right child.
There are multiple test cases. For each test case, this method will be called individually.

Output:
The function should return 1 if BST else return 0.

Constraints:
1 <=T<= 30
1 <= Number of nodes<= 100
1 <= Data of a node<= 1000

Example:
Input:
2
2
1 2 R 1 3 L
4
10 20 L 10 30 R 20 40 L 20 60 R

Output:
0
0

Both of the given trees are not BST.

There are two test casess.  First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 3 and right child of 1 is 2.   Second test case represents a tree with 4 edges and 5 nodes.

 CODE:

class GfG

{

    int isBST(Node root)  

    {

        // Your code 

        if(isBSTUtil(root, Integer.MIN_VALUE,

                               Integer.MAX_VALUE)){return 1;}

         return 0;                      

    }

    boolean isBSTUtil(Node node, int min, int max)

    {

        /* an empty tree is BST */

        if (node == null)

            return true;

        /* false if this node violates the min/max constraints */

        if (node.data < min || node.data > max)

            return false;

        /* otherwise check the subtrees recursively

        tightening the min/max constraints */

        // Allow only distinct values

        return (isBSTUtil(node.left, min, node.data-1) &&

                isBSTUtil(node.right, node.data+1, max));

    }

}

EXECUTION TIME: 0.1s