[Leetcode]Sum Root to Leaf Numbers

方法一：回溯法

1、解向量可以不用一个真的“向量”来存储，而可以用一个数来存放，这个数由路径上节点的值组成

2、每当一个解找到(到达叶子节点)，累加和

/** * 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 sumNumbers(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int ans = 0, sum = 0;        sumNumbers(root, ans, sum);        return sum;    }    void sumNumbers(TreeNode *root, int &ans, int &sum)    {        if(root == NULL) return;        ans = ans * 10 + root->val;        if(root->left == NULL && root->right == NULL)            sum += ans;        sumNumbers(root->left, ans, sum);        sumNumbers(root->right, ans, sum);        ans /= 10;    }};

方法二：递归

如果想用递归解决此题，必须把根节点的信息传递下去

class Solution {public:    int sumNumbers(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        return sumNumbers(root, 0);     }    int sumNumbers(TreeNode *root, int sum)    {        if(root == NULL) return 0;        sum = sum * 10 + root->val;        if(root->left == NULL && root->right == NULL) return sum;        return sumNumbers(root->left, sum) + sumNumbers(root->right, sum);    }};