Binary Tree Paths*
Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1
/ \
2 3
\
5
All root-to-leaf paths are:
["1->2->5", "1->3"]
Solution:
这是简单的DFS的遍历题目,由于需要不停的传递进入str变量,这个时候有两种做法,一种就是默认根节点有值,然后-> 指向子节点,另外一种就是我地下的做法。
class Solution(object):
def binaryTreePaths(self, root):
if root == None:
return []
result = []
self.dfs(root, result, '')
return result
def dfs(self, root, result, strs):
if root == None:
return
s = s + '->' + str(root.val) if s != '' else str(root.val)
if root.left == None and root.right == None:
result.append(strs)
if root.left:
self.dfs(root.left, result, strs)
if root.right:
self.dfs(root.right, result, strs)
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