标签：arch [] sub strong 二叉树 contain 完整 中序 .com
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a 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:
2 / 1 3Binary tree
[2,1,3], return true.
Example 2:
1 / 2 3Binary tree
[1,2,3], return false.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# 先将二叉搜索树进行中序遍历，接着判断是否满足规则
class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
self.ret = []
self.helper(root)
for i in xrange(1, len(self.ret)):
if self.ret[i] <= self.ret[i-1]:
return False
return True
def helper(self, node):
if node == None:
return
self.helper(node.left)
self.ret.append(node.val)
self.helper(node.right)
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
return self.helper(root, -float("inf"), float("inf"))
def helper(self, node, low, high):
if node == None:
return True
if node.val <= low or node.val >= high:
return False
return self.helper(node.left, low, node.val) and self.helper(node.right, node.val, high)
LeetCode 98. Validate Binary Search Tree
标签：arch [] sub strong 二叉树 contain 完整 中序 .com
原文地址：http://www.cnblogs.com/LiCheng-/p/6828976.html
