Leetcode #430: Flatten a Multilevel Doubly Linked List
In this guide, we solve Leetcode #430 Flatten a Multilevel Doubly Linked List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Linked List, Doubly-Linked List
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
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:
Python Solution
"""
# 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':
def preorder(pre, cur):
if cur is None:
return pre
cur.prev = pre
pre.next = cur
t = cur.next
tail = preorder(cur, cur.child)
cur.child = None
return preorder(tail, t)
if head is None:
return None
dummy = Node(0, None, head, None)
preorder(dummy, head)
dummy.next.prev = None
return dummy.next
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.