LeetCode - 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:

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

 

My Solution:

# 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:
        inorder_list = []

        def inorder_traverse(root):
            if root.left:
                inorder_traverse(root.left)
            inorder_list.append(root.val)
            if root.right:
                inorder_traverse(root.right)
        
        inorder_traverse(root)

        for i in range(len(inorder_list) - 1):
            if inorder_list[i + 1] <= inorder_list[i]:
                return False
        
        return True