Likou 98. Verifying Binary Search Trees

Topic source: 98. Verification of binary search tree
title:
Give you the root node root of a binary tree, and determine whether it is a valid binary search tree.
An efficient binary search tree is defined as follows:

• The left subtree of a node contains only numbers less than the current node.
• The right subtree of a node contains only numbers greater than the current node.
• All left and right subtrees must themselves be binary search trees.

Idea: In-order traversal
sees that the left node is smaller than the root node, and the right node is larger than the root node. I immediately thought that it might be related to in-order traversal. I looked at the answer and it was true. The idea is out, oh yeah
, the conditions given by the title are very similar to the rules of in-order traversal, so as long as the array obtained after in-order traversal is strictly in ascending order, that is, five repetitions, so at the end, a set function length judgment. The specific code is as follows:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        res = []
        def InOrderTraverse(r):
            if not r:
                return
            InOrderTraverse(r.left)
            res.append(r.val)
            InOrderTraverse(r.right)
        InOrderTraverse(root)
        if res == sorted(res) and len(res) == len(set(res)):
            return True
        else:
            return False

I don’t know if there is a better solution in the final judgment. I feel that the final running time and space are not very good. Partners with good ideas can discuss it together~
Reference link: Python Verification Binary Search Tree
LeetCode-Python- 98. Verify Binary Search Tree