LeetCode 430. Flatten a Multilevel Doubly Linked List

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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