1302.deepest-leaves-sum

Statement

简体中文English

Metadata

• Link: 层数最深叶子节点的和
• Difficulty: Medium
• Tag: 树 深度优先搜索 广度优先搜索 二叉树

给你一棵二叉树的根节点 root ，请你返回 层数最深的叶子节点的和 。

 

示例 1：

输入：root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
输出：15

示例 2：

输入：root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
输出：19

 

提示：

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

Metadata

• Link: Deepest Leaves Sum
• 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

Python 3

# 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

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最后更新: October 11, 2023