# 257.Binary Tree Paths

**257.Binary Tree Paths**

难度:Easy

> 给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

```
示例:

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3
```

方法：递归求解，直到叶子节点停止。

```
/**
 * 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) {
        class ALL_PATH {
            public:
            void path(TreeNode* root, string s, vector<string> &res)
            {
                if(!root->left && !root->right) 
                    res.push_back(s+to_string(root->val));
                if(root->left)
                    path(root->left, s+to_string(root->val)+"->", res);
                if(root->right)
                    path(root->right,s+to_string(root->val)+"->", res);
            }
        };
        vector<string> res;
        if(root)
        {
             ALL_PATH ap;
            ap.path(root,"",res);
        }     
        return res;        
        
    }
};
```


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