# 98 Validate Binary Search Tree

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:

``````  2
/ \
1   3

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

Example 2:

``````  5
/ \
1   4
/ \
3   6

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`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 `````` ``````# 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: def validate( node: Optional[TreeNode], low=-math.inf, high=math.inf ) -> bool: if not node: return True if node.val <= low or node.val >= high: return False return ( validate(node.left, low, node.val) and validate(node.right, node.val, high) ) return validate(root) ''' In order ''' class Solution: def isValidBST(self, root: TreeNode) -> bool: def inorder(root): if not root: return True if not inorder(root.left): return False if root.val <= self.prev: return False self.prev = root.val return inorder(root.right) self.prev = -math.inf return inorder(root)``````