Flatten a Multilevel Doubly Linked List

Tested Python solution for LeetCode 430 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 430, Medium. Topics: Linked List, Depth-First Search, Doubly-Linked List. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 430   # by problem number
lcpy gen -s flatten_a_multilevel_doubly_linked_list   # by problem name

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Problem

You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. 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.

Given the head of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr be a node with a child list. The nodes in the child list should appear after curr and before curr.next in the flattened list.

Return the head of the flattened list. The nodes in the list must have all of their child pointers set to null.

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Examples

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 shown.
After flattening the multilevel linked list it becomes:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:

Input: head = []
Output: []
Explanation: There could be empty list in the input.

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Constraints

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

How the multilevel linked list is represented in test cases:

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]

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from __future__ import annotations


class Node:
    def __init__(
        self,
        val: int = 0,
        prev: Node | None = None,
        next: Node | None = None,
        child: Node | None = None,
    ) -> None:
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child


class Solution:
    # Time: O(n) per level splice, O(n * depth) worst case
    # Space: O(1)
    def flatten(self, head: Node | None) -> Node | None:
        node = head
        while node is not None:
            if node.child is None:
                node = node.next
                continue
            child = node.child
            node.child = None
            nxt = node.next
            node.next = child
            child.prev = node
            tail = child
            while tail.next is not None:
                tail = tail.next
            tail.next = nxt
            if nxt is not None:
                nxt.prev = tail
            node = child
        return head

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Complexity

Time	Space
O(n) per level splice, O(n * depth) worst case	O(1)

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