LeetCode98:Validate Binary Search Tree
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:
Input: 2 / \ 1 3 Output: true
Example 2:
5 / \ 1 4 / \ 3 6 Output: false Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value is 5 but its right child's value is 4.
LeetCode:链接
题目要求去validate一个tree是不是Binary Search Tree,也就是所有left subtree的值都比现在node的值小,所有right subtree的值都比现在node的值大。
要注意的坑就是有可能是形如:
这就不是有效的binary search tree了,因为3虽然比8小,但是必须要比5大,所以在写代码的时候不能只比较当前父节点的大小,还要考虑之前所有的父节点情况。所以我们必须要有两个变量,lowerbound和upperbound。每次递归的时候更新这两个变量。
# 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
"""
lowerbound = float('-inf')
upperbound = float('inf')
return self.ValidBST(root, lowerbound, upperbound)
def ValidBST(self, root, lowerbound, upperbound):
if not root:
return True
if root.val <= lowerbound or root.val >= upperbound:
return False
return self.ValidBST(root.left, lowerbound, root.val) and self.ValidBST(root.right, root.val, upperbound)