题目如下:

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: [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.

Note:

  1. The number of nodes in the given tree is between 1 and 10^4.
  2. -10^5 <= node.val <= 10^5

解题思路:没什么好说的,依次计算出每一层的值,求出最大即可。

代码如下:

# 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):
dic = {}
def recursive(self,node,level):
self.dic[level] = self.dic.setdefault(level,0) + node.val
if node.left != None:
self.recursive(node.left,level+1)
if node.right != None:
self.recursive(node.right,level+1) def maxLevelSum(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.dic = {}
self.recursive(root,1)
max_val = 0
res = 0
for key,val in self.dic.iteritems():
if max_val < val:
max_val = val
res = key
return res

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