# 1302.deepest-leaves-sum

## Statement

• Difficulty: Medium
• Tag: `树` `深度优先搜索` `广度优先搜索` `二叉树`

``````输入：root = [1,2,3,4,5,null,6,7,null,null,null,null,8]

``````

``````输入：root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]

``````

• 树中节点数目在范围 `[1, 104]` 之间。
• `1 <= Node.val <= 100`

• Difficulty: Medium
• Tag: `Tree` `Depth-First Search` `Breadth-First Search` `Binary Tree`

Given the `root` of a binary tree, return the sum of values of its deepest leaves.

Example 1:

``````Input: root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
Output: 15
``````

Example 2:

``````Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
Output: 19
``````

Constraints:

• The number of nodes in the tree is in the range `[1, 104]`.
• `1 <= Node.val <= 100`

## 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
from typing import Optional

class Solution:
def __init__(self):
self.max_dep = 0
self.sum = 0

def dfs(self, rt, d):
if not rt:
return
d += 1

if d > self.max_dep:
self.sum = rt.val
self.max_dep = d
elif d == self.max_dep:
self.sum += rt.val

self.dfs(rt.left, d)
self.dfs(rt.right, d)

def deepestLeavesSum(self, root: Optional[TreeNode]) -> int:
self.dfs(root, 0)
return self.sum
``````