[leetcode] 1161. Maximum Level Sum of a Binary Tree

Description

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, 104].
• -105 <= Node.val <= 105

分析

题目的意思是：给你定二叉树，求每层和的最大值所在的层，根结点为第一层。这道题我一开始的想法就是用队列，然后用max_l记录最大值所在的层，max_v记录当前遍历的层的最大值。然后用队列进行层序遍历就行了。我看了一下其他人的解法，好像有跟我一样的哈哈哈。

代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxLevelSum(self, root: TreeNode) -> int:
        q=collections.deque()
        q.append(root)
        max_l=1
        max_v=root.val
        l=1
        while(q):
            n=len(q)
            cur_v=0
            for i in range(n):
                node=q.popleft()
                if(node.left):
                    q.append(node.left)
                if(node.right):
                    q.append(node.right)
                cur_v+=node.val
            if(cur_v>max_v):
                max_v=cur_v
                max_l=l
            l+=1
        return max_l