[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);
}
};