sum root to leafs nums

Given a binary tree containing digits from 0-9 only,
each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which
represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1   / \  2   3

The root-to-leaf path 1->2 represents
the number 12.

The root-to-leaf path 1->3 represents
the number 13.

Return the sum = 12 + 13 = 25.

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int sum;vector<int> tempNum;int toInt(vector<int>& num){        if(num.size() == 0)            return 0;int ret = 0;for(int i = 0; i < num.size(); i ++)ret = ret * 10 + num[i];return ret;}void dfs(TreeNode* root){if(root == NULL){            return;}        tempNum.push_back(root->val);        if(root->left == NULL && root->right == NULL)        {            sum += toInt(tempNum);        }dfs(root->left);dfs(root->right);tempNum.pop_back();}int sumNumbers(TreeNode *root) {// Start typing your C/C++ solution below// DO NOT write int main() functionsum = 0;        tempNum.clear();dfs(root);return sum;}};

只有葉子節點才可加，要判定是否為葉子節點