LeetCode | 0098. Validate Binary Search Tree验证二叉搜索树【Python】
LeetCode 0098. Validate Binary Search Tree验证二叉搜索树【Medium】【Python】【二叉树】
Problem
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 3
Input: [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: [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.
问题
给定一个二叉树,判断其是否是一个有效的二叉搜索树。
假设一个二叉搜索树具有如下特征:
- 节点的左子树只包含小于当前节点的数。
- 节点的右子树只包含大于当前节点的数。
- 所有左子树和右子树自身必须也是二叉搜索树。
示例 1:
输入:
2
/ \
1 3
输出: true
示例 2:
输入:
5
/ \
1 4
/ \
3 6
输出: false
解释: 输入为: [5,1,4,null,null,3,6]。
根节点的值为 5 ,但是其右子节点值为 4 。
思路
二叉树
root 不只需要和左右子节点比较,而是和整个左右子树所有节点比较。
Python3代码
# 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:
return self.isValid(root, None, None)
def isValid(self, root: TreeNode, min_: TreeNode, max_: TreeNode):
if not root:
return True
if min_ != None and root.val <= min_.val:
return False
if max_ != None and root.val >= max_.val:
return False
return self.isValid(root.left, min_, root) and self.isValid(root.right, root, max_)
GitHub链接
最怕一生碌碌无为,还说平凡难能可贵。