I found an interesting problem on https://leetcode.com/problems/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

I tried it in Python, but could not do it. Can you solve it?
# 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:
      if not root: return 
      st=[[root,0]]
      ll=[]
      while st:
        r,l=st.pop(0)
        if l==len(ll) : ll.append([])
        ll[l].append(r.val)
        if r.left: st.append([r.left,l+1])
        if r.right:
          st.append([r.right,l+1])
      #print (ll,l)
      m=-1000
      re=0
      for i in range(l+1):
        n=sum(ll[i])
        if m<n:
          m=n
          re=i
      return re+1
        
      
      
      \"\"\"
        if not root:
          return []
        queue,res=[[root, 0]],[]
        while queue:
          #print (queue[0][0].val)
          node,level = queue.pop(0)
          
          if level == len(res):
            res.append([])
        
          res[level].append(node.val)
          if node.left:
            queue.append((node.left, level + 1))
          if node.right:
            queue.append((node.right, level + 1))
          
        return res[::-1]
      \"\"\"