Binary Tree Paths
Created: September 13, 2018 by [lek-tin]
Last updated: September 13, 2018
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
Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def binaryTreePaths(self, root):
"""
:type root: TreeNode
:rtype: List[str]
"""
res = []
def traverse(root, path):
if root == None:
return
path += str(root.val) + "->"
if root.left == None and root.right == None:
res.append("".join(path[:-2]))
return
else:
traverse(root.left, list(path))
traverse(root.right, list(path))
traverse(root, "")
return res