# LeetCode 430. Flatten a Multilevel Doubly Linked List

## 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:

Example 2:

```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--NULL
The 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

class Node:
def __init__(self, val, prev, next, child):
self.val = val
self.prev = prev
self.next = next
self.child = child

class Solution:
def flatten(self, head: 'Node') -> 'Node':
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

```