LeetCode 430. Flatten a Multilevel Doubly Linked List

题目

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:

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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:

https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/


After flattening the multilevel linked list it becomes:

https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/

Example 2:

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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:

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Input: head = []
Output: []

How multilevel linked list is represented in test case:

We use the multilevel linked list from Example 1 above:

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

The serialization of each level is as follows:

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[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:

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[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:

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[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

思路

DFS.

代码

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"""
# 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):
def flatten(self, head):
"""
:type head: Node
:rtype: Node
"""
if not head: return None
node = head
while node:
node_next = node.next
if node.child:
flattened = self.flatten(node.child)
node.child = None
nextNode = self.appendToList(node, flattened)
node = nextNode
else:
node = node.next
return head


def appendToList(self, node, listToAppendHead):
next_node = node.next
node.next = listToAppendHead
listToAppendHead.prev = node
while node.next:
node = node.next
node.next = next_node
if next_node:
next_node.prev = node
return next_node