## 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]


Constraints:

• Number of Nodes will not exceed 1000.
• 1 <= Node.val <= 10^5

## Solutions

### 1. Coding Skill

主要考验的是编程能力，看能不能够 bug free 了！

# Time: O(n)
# Space: O(1)
"""
# Definition for a Node.
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 cur:
# CASE 1: if no child, proceed
if not cur.child:
cur = cur.next
continue
# CASE 2: got child, find the tail of the child and link it to cur.next
tmp = cur.child
# find the tail of the child
while tmp.next:
tmp = tmp.next
# connect tail with cur.next, if it is not None
tmp.next = cur.next
if cur.next:
cur.next.prev = tmp
cur.next = cur.child
cur.child.prev = cur
cur.child = None
# 22/22 cases passed (36 ms)
# Your runtime beats 72.4 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)


### 2. Stack + two pointers

总有一个先处理 child 后处理 next 的操作，就可以用 stack 来存。

class Solution:
def flatten(self, head: 'Node') -> 'Node':
prev = Node(0)
while stack:
node = stack.pop()
node.prev = prev
prev.next = node
prev = node
if node.next:
stack.append(node.next)
if node.child:
stack.append(node.child)
node.child = None