Problem:

Find the sum of all left leaves in a given binary tree.

Summary:

求左子叶之和。

Analysis：

1. 求左子叶之和，即需要遍历整棵二叉树，常规的方法则为递归。首先我想到的是调用子函数，在子函数中递归，每递归至一个新的节点，首先传入bool类型标记是根节点的左子树还是右子树。若为左子树，且为叶子节点，则将当前节点的value计入和中。

1 /** 2  * Definition for a binary tree node. 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     int sumOfLeftLeaves(TreeNode* root) {13         if (root == NULL) {14             return 0;15         }16         17         int sum = 0;18         calLeftLeaves(root->left, true, sum);19         calLeftLeaves(root->right, false, sum);20         21         return sum;22     }23 24 private:25     void calLeftLeaves(TreeNode* root, bool left, int& sum) {26         if (root == NULL) {27             return;28         }29         30         if (left && (root->left == NULL) && (root->right == NULL)) {31             sum += root->val;32         }33         34         calLeftLeaves(root->left, true, sum);35         calLeftLeaves(root->right, false, sum);36     }37 };

 2. 进一步简化代码，试图不掉用子函数，在主函数中进行递归操作。

1 /** 2  * Definition for a binary tree node. 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     int sumOfLeftLeaves(TreeNode* root) {13         if (root == NULL) {14             return 0;15         }16         17         if (root->left && root->left->left == NULL && root->left->right == NULL) {18             return root->left->val + sumOfLeftLeaves(root->right);19         }20         21         return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);22     }23 };