problem

 solution１:recursive递归方法。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void binaryTreePaths(vector
      
       & result, TreeNode* root, string str)//取地址符的用法.    {        if(!(root->left) && !(root->right))         {            result.push_back(str);            return;        }        if(root->left)             binaryTreePaths(result,root->left, str+"->"+to_string(root->left->val));//        if(root->right)             binaryTreePaths(result, root->right, str+"->"+to_string(root->right->val));    }    vector
       
         binaryTreePaths(TreeNode* root) {        vector
        
          result;        if(!root) return result;//        binaryTreePaths(result, root, to_string(root->val));        return result;            }};
        
       
      

solution2:DFS

 

solution3:BFS

 

 

 

参考

1. ;

2. ;

完