-
Notifications
You must be signed in to change notification settings - Fork 3
/
Copy path1161.maximum-level-sum-of-a-binary-tree.py
57 lines (39 loc) · 1.24 KB
/
1161.maximum-level-sum-of-a-binary-tree.py
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
from lc import *
class Solution:
def maxLevelSum(self, r: TreeNode) -> int:
c=Counter();f=lambda n,d:n and(c.update({d:n.val}),f(n.left,d+1),f(n.right,d+1));f(r,1);return max(c,key=c.get)
test('''
1161. Maximum Level Sum of a Binary Tree
Medium
2014
71
Add to List
Share
Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.
Return the smallest level x such that the sum of all the values of nodes at level x is maximal.
Example 1:
Input: root = [1,7,0,7,-8,null,null]
Output: 2
Explanation:
Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.
Example 2:
Input: root = [989,null,10250,98693,-89388,null,null,null,-32127]
Output: 2
Constraints:
The number of nodes in the tree is in the range [1, 10^4].
-10^5 <= Node.val <= 10^5
Accepted
132,994
Submissions
201,317
Seen this question in a real interview before?
Yes
No
Calculate the sum for each level then find the level with the maximum sum.
How can you traverse the tree ?
How can you sum up the values for every level ?
Use DFS or BFS to traverse the tree keeping the level of each node, and sum up those values with a map or a frequency array.
''')