98. Validate Binary Search Tree

Problem

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.

A subtree of treeName is a tree consisting of a node in treeName and all of its descendants.

https://leetcode.com/problems/validate-binary-search-tree/

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

Test Cases

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:

solution_test.py下载完整源码
py
import pytest

import os
import sys
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
from _utils.binary_tree import build_tree
from solution import Solution

null = None


@pytest.mark.parametrize('root, expected', [
    ([2,1,3], True),
    ([5,1,4,null,null,3,6], False),

    ([2,2,2], False),
    ([5,4,6,null,null,3,7], False),
])
class Test:
    def test_solution(self, root, expected):
        sol = Solution()
        assert sol.isValidBST(build_tree(root)) == expected

Thoughts

直接按照中序（in-order，LNR）遍历二叉树。如果是 BST，会得到一个严格递增的序列。遍历的时候拿当前节点和前一个节点比较，如果 key 不是严格增大的就不是 BST。

可以用栈和循环替代递归。

Code

solution.py下载完整源码
py
from typing import Optional


# 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:
        max_val = None
        stack = []
        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                if max_val is not None and root.val <= max_val:
                    return False
                max_val = root.val
                root = root.right

        return True

#Problem

#Test Cases

#Thoughts

#Code

Problem

Test Cases

Thoughts

Code