動機

其實有level也可以用dfs的

Problem

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,Leetcode,null]Output: 2Explanation: 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

Sol

class Solution:
    def maxLevelSum(self, root: Optional[TreeNode]) -> int:
        sums = defaultdict(int)
        def dfs(root, dep):
            if root:
                sums[dep] += root.val
                dfs(root.left, dep+1)
                dfs(root.right, dep+1)
        dfs(root, 1)
        return max(sums.items(), key=lambda p: p[1])[0]