二叉树从根开始每个结点存放一个0~9的数字，计算从根开始到叶节点生成的所有数字之和。从根开始搜索，到一个叶节点，将string转换加到结果里面就行。

class Solution {
public:
    int ConvertNumber(string &tmpNumber) {
        int number = 0;
        for(int i = 0;i < tmpNumber.length();i++) {
            number += (tmpNumber[tmpNumber.length() - 1 - i] - '0')*pow(10.0,i);
        }
        return number;
    }
    void AddNextLevel(TreeNode *node,int &sum,string tmpNumber) {
        tmpNumber.push_back(node->val + '0' - 0);
        if(node->left == NULL && node->right == NULL) {
            sum += ConvertNumber(tmpNumber);
            return ;
        }
        if(node->left != NULL) {
            AddNextLevel(node->left,sum,tmpNumber);
        }
        if(node->right != NULL) {
            AddNextLevel(node->right,sum,tmpNumber);
        }
    }
    int sumNumbers(TreeNode *root) {
        int sum = 0;
        if(root == NULL)return sum;
        string tmpNumber = "";
        AddNextLevel(root,sum,tmpNumber);
        return sum;
    }
};