大型# 1. 题目 #### [404. 左叶子之和](https://leetcode-cn.com/problems/sum-of-left-leaves/) 难度简单390 给定二叉树的根节点 `root` ,返回所有左叶子之和 **示例 1:** ![img](https://assets.leetcode.com/uploads/2021/04/08/leftsum-tree.jpg) ``` 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` # 2. 题解 # 3. code ```c++ class Solution { public: int sumOfLeftLeaves(TreeNode* root) { if (!root) return 0; int sum = 0; queue<TreeNode*> q; q.push(root); while (!q.empty()) { TreeNode* node = q.front(); q.pop(); if (node->left) { if (!node->left->left && !node->left->right) { sum += node->left->val; } else { q.push(node->left); } } if (node->right) { q.push(node->right); } } return sum; } }; ``` # 4. 心得

Rita Nazareth