城堡# 1. 題目 #### [404. 左葉子之和](https://leetcode-cn.com/problems/sum-of-left-leaves/) 難度簡單390 給定二叉樹的根節點 `root` ,返回所有左葉子之和 **示例 1:** ![img](https://assets.leetcode.com/uploads/2021/04/08/leftsum-tree.jpg) ``` 輸入: root = [3,9,20,null,null,15,7] 輸出: 24 解釋: 在這個二叉樹中,有兩個左葉子,分別是 9 和 15,所以返回 24 ``` **示例 2:** ``` 輸入: root = [1] 輸出: 0 ``` **提示:** - 節點數在 `[1, 1000]` 範圍內 - `-1000 <= Node.val <= 1000` # 2. 題解 # 3. code ```c++ class Solution { public: int sumOfLeftLeaves(TreeNode* root) { if (!root) return 0; queue<TreeNode*> q; q.push(root); int ans = 0; while (!q.empty()) { TreeNode* node = q.front(); q.pop(); if (node->left) { if (!node->left->left && !node->left->right) { ans += node->left->val; } else { q.push(node->left); } } if (node->right) { q.push(node->right); } } return ans; } }; ``` # 4. 心得

Hema Parmar, Nishant Kumar