palindrome partitioning

vector<vector<string>> partition(string s){
    vector<vector<string> >res;
    vector<string>path;
    
    dfs(s,path,res);
    
    return res;
}
private:
void dfs(string s, vector<string>& path, vector<vector<string> >& res){
    if(s.size() == 0){
        res.push_back(path);
        return;
    }
    
    for(int i=0; i<s.size(); i++){
        int left = 0, right = i;
        
        while(left < right){
            if(s[left] == s[right]){
                left++;
                right--;
            }
            else
                break;
        }
        
        if(left >= right){//有回文，前面
            path.push_back(s.substr(0,i+1));//i+1个字符
            dfs(s.substr(i+1),path,res);
            path.pop_back();
        }
    }
}