# leetcode\_257

## 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
```

## Solutions

1. **recursive(dfs)**

```python
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root: return []
        return [root.val + '->' + subpath
                for kid in (root.left, root.right) if kid
                for subpath in self.binaryTreePaths(kid)
                ] or [ str(root.val) ]
```

```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> path;
    vector<string> res;
    void dfs(TreeNode * root) {
        if (!root->left && !root->right) {
            string s;
            for (auto & subs : path)
                s += subs + "->";
            res.push_back(s + to_string(root->val));
            return;
        }
        path.push_back(to_string(root->val));
        if (root->left) dfs(root->left);
        if (root->right) dfs(root->right);
        path.pop_back();
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        if (root) dfs(root);
        return res;
    }
};
```