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LC1161 - Maximum Level Sum of a Binary Tree

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Datastructure and Algorithm: DAY 61. Leetcode: 226/2726

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:

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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:

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Input: root = [989,null,10250,98693,-89388,null,null,null,-32127]
Output: 2

Constraints:

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The number of nodes in the tree is in the range [1, 104].
-105 <= Node.val <= 105

Solution

BFS counts the sum of each layer, and returns the one with the largest sum and the smallest number of layers.

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# 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
from collections import deque

class Solution:
    def maxLevelSum(self, root: Optional[TreeNode]) -> int:
        queue = deque([root])
        maxi = float('-inf')
        level = 1
        ans = 0
        while queue:
            l = len(queue)
            summ = 0
            for _ in range(l):
                cur = queue.popleft()
                summ += cur.val
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            if summ > maxi:
                maxi = summ
                ans = level
            level+=1
        return ans