给一棵二叉树，找出从根节点到叶子节点的所有路径。
样例 1:
输入：{1,2,3,#,5}
输出：["1->2->5","1->3"]
解释：
1
/ 2 3
5
样例 2:
输入：{1,2}
输出：["1->2"]
解释：
1
/
2
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: the root of the binary tree
@return: all root-to-leaf paths
"""
def binaryTreePaths(self, root):
# write your code here
path = []
result = []
self.dfs(root, path, result)
return result
def dfs(self, root, path, result):
if not root:
return
path.append(str(root.val))
if root.left is None and root.right is None:
result.append("->".join(path))
else:
self.dfs(root.left, path, result)
self.dfs(root.right, path, result)
path.pop()
还有使用分治实现的，jiuzhang给的：
class Solution:
"""
@param root: the root of the binary tree
@return: all root-to-leaf paths
"""
def binaryTreePaths(self, root):
if root is None:
return []
if root.left is None and root.right is None:
return [str(root.val)]
leftPaths = self.binaryTreePaths(root.left)
rightPaths = self.binaryTreePaths(root.right)
paths = []
for path in leftPaths + rightPaths:
paths.append(str(root.val) + ‘->‘ + path)
return paths
原文：https://www.cnblogs.com/bonelee/p/11610113.html