257. Binary Tree Paths

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

# 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 root is None:
            return None
        self.res = []
        self.helper(root, [])
        out = []
        for l in self.res:
            out.append("->".join(l))
        return out
    def helper(self, root, tmpl):
        if root.left is None and root.right is None:
            self.res.append(tmpl + [str(root.val)])       
        if root.left:
            self.helper(root.left, tmpl + [str(root.val)])
        if root.right:
            self.helper(root.right, tmpl + [str(root.val)])