In this post, we are going to solve the 98. Validate Binary Search Tree problem of Leetcode. This problem 98. Validate Binary Search Tree is a Leetcode medium level problem. Let’s see the code, 98. Validate Binary Search Tree – Leetcode Solution.

Problem

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

A valid 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 :

Input: root = [2,1,3]
Output: true

Example 2 :

Input: root = [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.

Constraints

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

Now, let’s see the code of 98. Validate Binary Search Tree – Leetcode Solution.

Validate Binary Search Tree – Leetcode Solution

98. Validate Binary Search Tree – Solution in Java

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

98. Validate Binary Search Tree – Solution in C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        long pre = LONG_MIN;
        stack<TreeNode*> todo;
        while (root || !todo.empty()) {
            while (root) {
                todo.push(root);
                root = root -> left;
            }
            root = todo.top();
            todo.pop();
            if (root -> val <= pre) {
                return false;
            }
            pre = root -> val;
            root = root -> right;
        }
        return true;
    }
};

98. Validate Binary Search Tree – Solution in Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isValidBST(self, root):
        pre, stack = None, []
        while True:
            while root:
                stack.append(root)
                root = root.left
            if not stack:
                return True
            node = stack.pop()
            if pre and pre.val >= node.val:
                return False
            pre = node
            root = node.right
        

Note: This problem 98. Validate Binary Search Tree is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.