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# Validate Binary Search Tree LeetCode Solution

## Problem – Validate Binary Search Tree LeetCode Solution

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

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

### Validate Binary Search Tree LeetCode Solution in Java

public class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}

public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
if (root == null) return true;
if (root.val >= maxVal || root.val <= minVal) return false;
return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
}
}

### Validate Binary Search Tree LeetCode Solution in C++

class Solution {
public:
bool isValidBST(TreeNode* root) {
TreeNode* prev = NULL;
return validate(root, prev);
}
bool validate(TreeNode* node, TreeNode* &prev) {
if (node == NULL) return true;
if (!validate(node->left, prev)) return false;
if (prev != NULL && prev->val >= node->val) return false;
prev = node;
return validate(node->right, prev);
}
};

### Validate Binary Search Tree LeetCode Solution in Python

class Solution(object):
def isValidBST(self, root, lessThan = float('inf'), largerThan = float('-inf')):
if not root:
return True
if root.val <= largerThan or root.val >= lessThan:
return False
return self.isValidBST(root.left, min(lessThan, root.val), largerThan) and \
self.isValidBST(root.right, lessThan, max(root.val, largerThan))
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