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Description

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

Example 1:

Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]Output: [1,2,3,7,8,11,12,9,10,4,5,6]Explanation:The multilevel linked list in the input is as follows:

After flattening the multilevel linked list it becomes:

Input: head = [1,2,null,3]Output: [1,3,2]Explanation:

The input multilevel linked list is as follows:  1---2---NULL  |  3---NULL

Example 3:

Input: head = []Output: []

How multilevel linked list is represented in test case:

We use the multilevel linked list from Example 1 above: 1---2---3---4---5---6--NULL         |         7---8---9---10--NULL             |             11--12--NULLThe serialization of each level is as follows:[1,2,3,4,5,6,null][7,8,9,10,null][11,12,null]To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:[1,2,3,4,5,6,null][null,null,7,8,9,10,null][null,11,12,null]Merging the serialization of each level and removing trailing nulls we obtain:[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]

思路

• 使用循环，从头开始遍历，将子链看成一个整体，将整体向上移动一层。
• 继续向后遍历，如果遇到子链，将其看作一层，将其整体向上移动，如此循环下去，直至结束。

# -*- coding: utf-8 -*-# @Author:             何睿# @Create Date:        2019-12-29 10:57:03# @Last Modified by:   何睿# @Last Modified time: 2019-12-29 11:05:17class Node:    def __init__(self, val, prev, next, child):        self.val = val        self.prev = prev        self.next = next        self.child = childclass Solution:    def flatten(self, head: 'Node') -> 'Node':        current = head        while current:            if current.child:                _next = current.next  # 当前节点的后一个节点                last = current.child  # 当前节点的自节点                while last.next:                    last = last.next  # 如果有子节点，找到子节点的最后一个节点                current.next = current.child                current.next.prev = current  # 将自节点向上提高一层                current.child = None                last.next = _next                if _next:                    _next.prev = last            current = current.next        return head

源代码文件在 这里 。

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