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leetcode-98-验证二叉搜索树

时间:2019-07-14 16:54:05      阅读:108      评论:0      收藏:0      [点我收藏+]

题目描述:

技术分享图片

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

 

leetcode-98-验证二叉搜索树

原文:https://www.cnblogs.com/oldby/p/11184220.html

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