1 Star 0 Fork 0

wd6/LeetCode-1

加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
文件
克隆/下载
flatten-binary-tree-to-linked-list.py 1.84 KB
一键复制 编辑 原始数据 按行查看 历史
# Time: O(n)
# Space: O(h), h is height of binary tree
#
# Given a binary tree, flatten it to a linked list in-place.
#
# For example,
# Given
#
# 1
# / \
# 2 5
# / \ \
# 3 4 6
# The flattened tree should look like:
# 1
# \
# 2
# \
# 3
# \
# 4
# \
# 5
# \
# 6
#
# Definition for a binary tree node
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# @param root, a tree node
# @return nothing, do it in place
def flatten(self, root):
return self.flattenRecu(root, None)
def flattenRecu(self, root, list_head):
if root != None:
list_head = self.flattenRecu(root.right, list_head)
list_head = self.flattenRecu(root.left, list_head)
root.right = list_head
root.left = None
return root
else:
return list_head
class Solution2:
list_head = None
# @param root, a tree node
# @return nothing, do it in place
def flatten(self, root):
if root != None:
self.flatten(root.right)
self.flatten(root.left)
root.right = self.list_head
root.left = None
self.list_head = root
return root
if __name__ == "__main__":
root = TreeNode(1)
root.left = TreeNode(2)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right = TreeNode(5)
root.right.right = TreeNode(6)
result = Solution().flatten(root)
print result.val
print result.right.val
print result.right.right.val
print result.right.right.right.val
print result.right.right.right.right.val
print result.right.right.right.right.right.val
Loading...
马建仓 AI 助手
尝试更多
代码解读
代码找茬
代码优化
Python
1
https://gitee.com/wd6/LeetCode-1.git
git@gitee.com:wd6/LeetCode-1.git
wd6
LeetCode-1
LeetCode-1
master

搜索帮助