-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy path110BalancedBinaryTree.py
More file actions
75 lines (65 loc) · 1.47 KB
/
110BalancedBinaryTree.py
File metadata and controls
75 lines (65 loc) · 1.47 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
"""
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
"""
"""
Comments
"""
"""
My
"""
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
def balDepth(node):
if node is None:
return 0
left = balDepth(node.left)
right = balDepth(node.right)
if left == -1 or right == -1 or abs(left-right) > 1:
return -1
return max(left,right)+1
return balDepth(root) != -1
"""
Fast
"""
class Solution:
def isBalanced(self, root: 'TreeNode') -> 'bool':
r = True
def f(node):
nonlocal r
if not node:
return 0
dl = f(node.left)
if not r:
return 0
dr = f(node.right)
if abs(dl - dr) > 1:
r = False
return max(dl, dr) + 1
f(root)
return r