#+PROPERTY: STUDY_DECK_02
* TODO 0098. Validate Binary Search Tree :medium:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0098. Validate Binary Search Tree][0098. Validate Binary Search Tree]]
:END:

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 *strictly less than* the node's key.

 - The right subtree of a node contains only nodes with keys *strictly greater than* the node's key.

 - Both the left and right subtrees must also be binary search trees.

*Example 1:*


#+begin_src
Input: root = [2,1,3]
Output: true
#+end_src


*Example 2:*


#+begin_src
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.
#+end_src


*Constraints:*

 - The number of nodes in the tree is in the range ~[1, 10^{4}]~.

 - ~-2^{31} <= Node.val <= 2^{31} - 1~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isValidBST(self, root: Optional[TreeNode]) -> bool:
#+end_src

** TODO C++
#+begin_src cpp
/**
 * 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) {
        
    }
};
#+end_src