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+# [404. Sum of Left Leaves](https://leetcode.com/problems/sum-of-left-leaves/description/)
+# 思路
+计算一棵二叉树所有左叶子的值之和。
+## 思路一
+设置一个初始为0的全局变量res，遍历一遍树，判断某节点的左子树是否是叶子，若是，则将这个叶子的值加至res，遍历完后返回res即可。
+
+## 思路二
+直接递归计算结果res。因为某棵树的res等于其左子树的res加上右子树的res。
+
+# C++
+## 思路一
+``` C++
+class Solution {
+private:
+    int res = 0;
+    bool is_leaf(TreeNode *p){ // 判断是否是叶子
+        if(p == NULL) return false;
+        return (p -> left == NULL && p -> right == NULL);
+    }
+    void find_left_leaf(TreeNode *root){
+        if(root == NULL || is_leaf(root)) return; 
+        
+        if(is_leaf(root -> left)) res += root -> left -> val; // 左子树是叶子，加至res
+        else find_left_leaf(root -> left);
+        find_left_leaf(root -> right);
+    }
+public:
+    int sumOfLeftLeaves(TreeNode* root) {
+        find_left_leaf(root);
+        return res;
+    }
+};
+```
+## 思路二
+``` C++
+class Solution {
+private:
+    bool is_leaf(TreeNode *p){ // 判断是否是叶子
+        if(p == NULL) return false;
+        return (p -> left == NULL && p -> right == NULL);
+    }
+public:
+    int sumOfLeftLeaves(TreeNode* root) {
+        if(root == NULL || is_leaf(root)) return 0;
+        if(is_leaf(root -> left)) return (root -> left -> val) + sumOfLeftLeaves(root -> right);
+        else return sumOfLeftLeaves(root -> left) + sumOfLeftLeaves(root -> right);        
+    }
+};
+```