Validate Binary Search Tree - The Coding Shala

Last Updated: 27-Jan-2021

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 In this post, we will learn how to Validate Binary Search Tree and will implement its solution in Java.

Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

Example 1:

    2

   / \

  1   3

Input: [2,1,3]

Output: true

Example 2:

    5

   / \

  1   4

      / \

    3   6

Input: [5,1,4,null,null,3,6]

Output: false

Explanation: The root node's value is 5 but its right child's value is 4.

Validate Binary Search Tree Java Program

Approach 1

Using recursion and inorder traversal. We can get the list of nodes in In-Order and InOrder is always in ascending order of value.

Java Program: 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inOrder(TreeNode root){
        List<Integer> In = new ArrayList<Integer>();
        if(root == null) return In;
        In.addAll(inOrder(root.left));
        In.add(root.val);
        In.addAll(inOrder(root.right));
        return In;
    }
    
    public boolean isValidBST(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        list = inOrder(root);
        for(int i=0; i<list.size()-1;i++){
            if(list.get(i)>=list.get(i+1)) return false;
        }
        return true;
    }
}

Approach 2

Iterative Solution.[Using Stack/dfs]

Java Program: 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null) return true;
        Stack<TreeNode> st = new Stack<TreeNode>();
        TreeNode check = null;
        while(root != null || !st.empty()){
            while(root != null){
                st.push(root);
                root = root.left;
            }
            root = st.pop();
            if(check != null && root.val <= check.val) return false;
            check = root;
            root = root.right;
        }
        return true;
    }
}

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