404.sum-of-left-leaves

Statement

简体中文English

Metadata

• Link: 左叶子之和
• Difficulty: Easy
• Tag: 树 深度优先搜索 广度优先搜索 二叉树

给定二叉树的根节点 root ，返回所有左叶子之和。

 

示例 1：

输入: root = [3,9,20,null,null,15,7] 
输出: 24 
解释: 在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24

示例 2:

输入: root = [1]
输出: 0

 

提示:

• 节点数在 [1, 1000] 范围内
• -1000 <= Node.val <= 1000

 

Metadata

• Link: Sum of Left Leaves
• Difficulty: Easy
• Tag: Tree Depth-First Search Breadth-First Search Binary Tree

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 24
Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:

Input: root = [1]
Output: 0

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -1000 <= Node.val <= 1000

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.res = 0

    def dfs(self, root: Optional[TreeNode], is_left: bool) -> None:
        if not root:
            return

        son = 0
        if root.left:
            son += 1
            self.dfs(root.left, True)

        if root.right:
            son += 1
            self.dfs(root.right, False)

        if son == 0 and is_left:
            self.res += root.val

    def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
        self.dfs(root, False)
        return self.res

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最后更新: October 11, 2023