"""
BFS遍历题,一遍AC
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.
"""
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None class Solution:
def maxLevelSum(self, root):
if root == None:
return 0
queue = []
res = []
queue.append(root)
while queue:
n = len(queue)
sum = 0
newqueue = []
for _ in range(n):
x = queue.pop(0)
sum += x.val
if x.left != None:
newqueue.append(x.left)
if x.right != None:
newqueue.append(x.right)
queue = newqueue
res.append(sum)
return res.index(max(res))+1

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