# LeetCode: 430. Flatten a Multilevel Doubly Linked List¶

## 1、题目描述¶

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:

Input:
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL

Output:
1-2-3-7-8-11-12-9-10-4-5-6-NULL


Explanation for the above example:

Given the following multilevel doubly linked list:

We should return the following flattened doubly linked list:

## 2、解题思路¶

​ 可以使用深度优先搜索，也可以不是用，首先来剖析一下问题

​ 这道题与前面114题很相似，如果当前的节点有子节点，该如何做呢？

​ 将剩下的节点，挂到子节点的末尾，然后，将当前节点的next指向子节点，子节点的prev指向当前节点即可，然后节点后移，继续判断

class Solution(object):
"""
:rtype: Node
"""

while temp.next:
temp = temp.next

return ans


​ 思路很简单，但是通过了19个测试用了，下一个超时了，因此，要考虑如何提高效率

​ 考虑在展开的过程中，不直接将下一个元素挂到子列的尾部，而是保存到栈中，等到遍历到最后的时候，就将栈中的元素取出来，接到尾部，然后继续判断即可

"""
# Definition for a Node.
class Node(object):
def __init__(self, val, prev, next, child):
self.val = val
self.prev = prev
self.next = next
self.child = child
"""
class Solution(object):
"""
:rtype: Node
"""
return ans

stack = []
prev = None
else:
temp = stack.pop()
prev.next = temp
temp.prev = prev

return ans


​ 这样，仅需要从头到尾遍历一遍即可得出结果