## DEV Community

Abhishek Chaudhary

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# Maximum Level Sum of a Binary Tree

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`

SOLUTION:

``````# 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 inorder(self, root, l):
if root:
self.inorder(root.left, l + 1)
self.levels[l] = self.levels.get(l, 0) + root.val
self.inorder(root.right, l + 1)

def maxLevelSum(self, root: Optional[TreeNode]) -> int:
self.levels = {}
self.inorder(root, 1)
return -max([(v, -k) for k, v in self.levels.items()])[1]
``````