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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

   /**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
bool isBst(TreeNode *root, int minv, int maxv)
{
 
   
    if(!root)
        return true;
    
    if(root->val > minv && root->val < maxv)
    {
        return isBst(root->left, minv, root->val)  &&
        isBst(root->right, root->val, maxv);
    }

    return false;
}

bool isValidBST(TreeNode *root) {
    return isBst(root, INT_MIN, INT_MAX);
}
};