# 257. Binary Tree Paths

## problem description

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

**Example:**

```
Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3
```

## algorithm thought

就是将DFS的路线标记出来，最后返回。直接使用DFS做即可

只是这题需要注意一些细节，不然容易过不了一些测试样例

## code

```cpp
/**
 * 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:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if(root==nullptr)
            return res;
        string path=to_string(root->val);
        if(root->left!=nullptr)
            help(res,root->left,path);
        path=to_string(root->val);
        if(root->right!=nullptr)
            help(res,root->right,path);
        if(root->left==nullptr&&root->right==nullptr){
            res.push_back(path);
        }
        return res;
    }
    void help(vector<string>&res,TreeNode* root,string&path){
        int sz=path.size();
        string number=to_string(root->val);
        path+="->";path+=number;
        if(root->left==nullptr&&root->right==nullptr){
            res.push_back(path);
            path.resize(sz);
            return;
        }
        if(root->left!=nullptr)
            help(res,root->left,path);
        if(root->right!=nullptr)
            help(res,root->right,path);
        path.resize(sz);
    }
};
```

## algorithm analysis

DFS将树遍历一遍，时间复杂度O(n)。


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