LeetCode 404 Sum of Left Leaves

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。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }
        
        int sum = 0;
        calLeftLeaves(root->left, true, sum);
        calLeftLeaves(root->right, false, sum);
        
        return sum;
    }

private:
    void calLeftLeaves(TreeNode* root, bool left, int& sum) {
        if (root == NULL) {
            return;
        }
        
        if (left && (root->left == NULL) && (root->right == NULL)) {
            sum += root->val;
        }
        
        calLeftLeaves(root->left, true, sum);
        calLeftLeaves(root->right, false, sum);
    }
};

 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。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }
        
        if (root->left && root->left->left == NULL && root->left->right == NULL) {
            return root->left->val + sumOfLeftLeaves(root->right);
        }
        
        return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
};