# 資訊

:::info
- Question: 404. Sum of Left Leaves
- From: Leetcode Daily Challenge 2024.0414
- Difficulty: Easy
:::

---

# 目錄

:::info
[TOC]
:::

---

# 題目

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:
:::success
* 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:
:::success
* Input: root = [1]
* Output: 0
:::

> Constraints:
:::success
* The number of nodes in the tree is in the range `[1, 1000]`.
* -1000 <= `Node.val` <= 1000
:::

---

# 解法

## 概念

這題就是簡單的 DFS 走訪過去就好，不過因為卡了一筆測資 `[1,2]` 才發現並不是所有「最後一片葉子」都不能算，而是要看是不是左邊再決定要不要算，因此這次 DFS 中多帶了一個 flag `left`，指出是不是在左邊，只有在左邊的才需要加到答案裡面

## 程式碼

# 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
class Solution:
    def __init__(self):
        self.sum = 0
    def dfs(self, root: TreeNode, left: bool ):
        if not root.left and not root.right and left == True:
            self.sum += root.val
        else:
            if root.left:
                self.dfs(root.left, True)
            if root.right:
                self.dfs(root.right, False)
    def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
        self.dfs(root, False)
        return self.sum

---

# 複雜度

## 時間複雜度

因為是看完整棵 tree，所以是 $O(n)$


## 空間複雜度

空間部分應該可以算 $O(1)$