Sum Root to Leaf Numbers - LintCode

描述
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.

A leaf is a node with no children.

样例
Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

思路

#ifndef C1353_H
#define C1353_H
#include<iostream>
using namespace std;
class TreeNode{
public:
    int val;
    TreeNode *left, *right;
    TreeNode(int val){
        this->val = val;
        this->left = this->right = NULL;
    }
};
class Solution {
public:
    /**
    * @param root: the root of the tree
    * @return: the total sum of all root-to-leaf numbers
    */
    int sumNumbers(TreeNode * root) {
        // write your code here
        if (!root)
            return 0;
        int num = 0;
        helper(root, num);
        return sum;
    }
    //计算节点root所表示的数字
    void helper(TreeNode *root, int &num)
    {
        if (!root)
            return;
        num = num*10 + root->val;
        //当root为叶子节点,将num的值加到sum上
        if (!root->left&&!root->right)
            sum += num;
        if (root->left)
        {
            helper(root->left, num);
            num -= root->left->val;
            num /= 10;
        }
        if (root->right)
        {
            helper(root->right, num);
            num -= root->right->val;
            num /= 10;
        }
    }
    int sum = 0;
};
#endif