Flatten a Multilevel Doubly Linked List

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:

分析

dfs返回首尾，记得如果有儿子，返回的tail是儿子的tail

"""
# 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':
        if not head:
            return head       
        def dfs(head):
            cur = head
            tail = None
            while cur:
                if cur.child:
                    h,e = dfs(cur.child)
                    cur.child = None
                    nxt = cur.next
                    cur.next = h
                    h.prev = cur
                    e.next = nxt
                    if nxt:                      
                        nxt.prev = e
                    cur = nxt
                    tail = e #tail有child的要指向e！！！！！
                else:
                    tail = cur  #否则指向最后一个item            
                    cur = cur.next

            return (head, tail)        


        h,e = dfs(head)
        return h