404.leetcode Sum of Left Leaves(easy)[二叉树 递归]

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

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

/**
 * 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:
    void getLeftSum(TreeNode* root,int flag,int &sum)
    {
        if(root->left != NULL)
        {
            flag = 1;
            getLeftSum(root->left,flag,sum);
        }
        if(root->right != NULL)
        { 
            flag = 0;
            getLeftSum(root->right,flag,sum);
        }
        if(root->left == NULL && root->right == NULL)
        {
            if(flag == 1)
               sum += root->val;
        }
        return;
    }
    int sumOfLeftLeaves(TreeNode* root) {
        if(root == NULL) return 0;
        int flag = 0;
        int sum = 0;
        getLeftSum(root,flag,sum);
        return sum;
    }
};