sum-root-to-leaf-numbers

Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path1->2->3which represents the number123.

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

For example,

1
/ \
2   3

The root-to-leaf path1->2represents the number12.
The root-to-leaf path1->3represents the number13.

Return the sum = 12 + 13 =25.

我的代码

**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void sum(TreeNode *root,vector<int>&path,int &res)
{
　　if(root==NULL)
　　　　　　return;
　　path.push_back(root->val);
　　if(root->left==NULL && root->right==NULL)
　　{
　　int sum=0;
　　for(int i=0;i<path.size();i++)
　　{
　　　　sum=sum*10+path[i];
　　}

　　res+=sum;
　　path.pop_back();
　　}else
　　{
　　　　sum(root->left,path,res);
　　　　sum(root->right,path,res);
　　　　path.pop_back();
　　}
}
int sumNumbers(TreeNode *root) {
　　vector<int>path;
　　int res=0;
　　sum(root,path,res);
　　return res;
}
};

别人的代码

int getsum(TreeNode *root,int num)
{
　　if(root==NULL)
　　　　　　return 0;
　　num=num*10+root->val;
　　if(root->left==NULL &&root->right==NULL)
　　　　　　return num;
　　return getsum(root->left,num)+getsum(root->right,num);
}
int sumNumbers(TreeNode *root) {

　　int sum=0;
　　sum=getsum(root,0);
　　return sum;