leetcode-98-验证二叉搜索树-优快云博客

博客介绍了解决某题目的三种方法，分别是递归、迭代和中序遍历，时间复杂度和空间复杂度均为O(n)，还给出了转载来源。

题目描述：

方法一：递归O(n) O(n)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        def isvalidBST(root,lower=float('-inf'),upper=float('inf')):
            if not root:
                return True
            val = root.val
            if val<=lower or val>=upper:
                return False
            if not isvalidBST(root.left,lower,val):
                return False
            if not isvalidBST(root.right,val,upper):
                return False
            return True
        
        return isvalidBST(root)

方法二：迭代O(n) O(n)

class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        if not root:
            return True
        stack = [(root.val,float('-inf'),float('inf'))]
        while stack:
            root, lower, upper = stack.pop()
            if not root:
                continue
            val = root.val if val <= lower or val >= upper: 
                return False 
            stack.append((root.right, val, upper)) 
            stack.append((root.left, lower, val))
        return True

方法三：中序遍历 O(n) O(n)

class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        stack, inorder = [], float('-inf')
        while stack or root:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            if root.val<=inorder:
                return False
            inorder = root.val
            root = root.right
        return True