[leetcode] 426. Convert Binary Search Tree to Sorted Doubly Linked List @ python

最新推荐文章于 2024-01-22 16:06:43 发布

原创最新推荐文章于 2024-01-22 16:06:43 发布 · 1.3k 阅读

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本文介绍了一种将二叉搜索树转换为排序双向链表的方法，使用递归方式构建链表，并确保时间复杂度为O(n)，空间复杂度为O(n)。通过定义辅助函数处理子链表连接，最终返回头结点。

原题

https://leetcode.com/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/

解法

使用递归, base case是如果node为空, 返回None. 定义helper函数, 使用头结点和尾节点, 将子链表的头结点和尾节点与母链表相连.
Time: O(v), v是二叉树的节点数目
Space: O(v)

代码

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right):
        self.val = val
        self.left = left
        self.right = right
"""
class Solution:
    def treeToDoublyList(self, root):
        """
        :type root: Node
        :rtype: Node
        """
        # base case
        if not root: return None
        head, tail = self.helper(root)
        return head
        
    def helper(self, root):
        '''construct a doubly-linked list, return the head and tail'''
        head, tail = root, root
        if root.left:
            left_head, left_tail = self.helper(root.left)
            left_tail.right = root
            root.left = left_tail
            head = left_head
        if root.right:
            right_head, right_tail = self.helper(root.right)
            right_head.left = root
            root.right = right_head
            tail = right_tail
            
        head.left = tail
        tail.right = head
        return head, tail