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]