-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy path110.平衡二叉树.py
More file actions
77 lines (74 loc) · 1.44 KB
/
110.平衡二叉树.py
File metadata and controls
77 lines (74 loc) · 1.44 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
#
# @lc app=leetcode.cn id=110 lang=python
#
# [110] 平衡二叉树
#
# https://leetcode-cn.com/problems/balanced-binary-tree/description/
#
# algorithms
# Easy (49.16%)
# Likes: 168
# Dislikes: 0
# Total Accepted: 33.4K
# Total Submissions: 67.9K
# Testcase Example: '[3,9,20,null,null,15,7]'
#
# 给定一个二叉树,判断它是否是高度平衡的二叉树。
#
# 本题中,一棵高度平衡二叉树定义为:
#
#
# 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1。
#
#
# 示例 1:
#
# 给定二叉树 [3,9,20,null,null,15,7]
#
# 3
# / \
# 9 20
# / \
# 15 7
#
# 返回 true 。
#
# 示例 2:
#
# 给定二叉树 [1,2,2,3,3,null,null,4,4]
#
# 1
# / \
# 2 2
# / \
# 3 3
# / \
# 4 4
#
#
# 返回 false 。
#
#
# @lc code=start
# 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 isBalanced(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
self.res=True
def depth(root):
if not root:return 0
left=depth(root.left)+1
right=depth(root.right)+1
if abs(left-right)>1:self.res=False
return max(left,right)
depth(root)
return self.res
# @lc code=end