Flatten Binary Tree to Linked List Leetcode Python

hyperbolechi

于 2015-01-25 10:25:58 发布

阅读量493

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： python leetcode Tree linklist

本文链接：https://blog.youkuaiyun.com/hyperbolechi/article/details/43112533

leetcode 专栏收录该内容

166 篇文章

订阅专栏

本文介绍了一种将二叉树通过递归方式展平为链表的方法。通过调整节点指针，使得每个节点的左子节点为空，右子节点成为下一个节点，最终形成链表结构。此算法的时间复杂度为O(n)，空间复杂度为log n。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
\

最近在刷面经，老看到考这题，于是又写了一遍。

we can use recursion to solve this problem. Define a prenode first, then in the recursion, everytime we put the left node as null and the right node as the new root.

remember we need to preserve the right root, otherwise the right root maybe replaced.

time complexity is O(n) space is logn

code is as follow.

# 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 solve(self,root):
        if root==None:
            return
        if self.prenode!=None:
            self.prenode.left=None
            self.prenode.right=root
        self.prenode=root
        left=root.left
        right=root.right
        self.solve(left)
        self.solve(right)
        return root
    def flatten(self, root):
        Solution.prenode=None
        self.solve(root)
        return root