1116 Maximum Level Sum Of A Binary Tree

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

# 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: Optional[TreeNode]) -> int:
        
        arr = [root]
        mx_sum = -100000
        mx_lvl = {"ans": 0}

        def recurse(arr, lvl, mx_sum, mx_lvl):

            if len(arr) == 0 :
                return
            curr_sum = 0
            curr_lvl = 0
            temp = []
            while(arr):
                node = arr.pop()
                if node.left:
                    temp.append(node.left)
                if node.right:
                    temp.append(node.right)
                curr_sum += node.val
            #print(curr_sum, mx_sum, lvl, mx_lvl)
            if curr_sum > mx_sum:
                mx_lvl["ans"] = lvl
                mx_sum = curr_sum
            recurse(temp, lvl + 1, mx_sum, mx_lvl)

        recurse(arr, 1, mx_sum, mx_lvl)
        return mx_lvl["ans"]