How do you solve Flatten a Multilevel Doubly Linked List on LeetCode?

Flatten a Multilevel Doubly Linked List (LeetCode #430) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: DFS is a powerful technique for tree-like structures.

What is the brute force solution for Flatten a Multilevel Doubly Linked List?

The brute force approach for Flatten a Multilevel Doubly Linked List is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we traverse the entire multilevel linked list and collect all nodes in a temporary list. Then, we reconstruct a single-level linked list from this temporary list. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Flatten a Multilevel Doubly Linked List?

Flatten a Multilevel Doubly Linked List uses the following concepts: Linked List, Depth-First Search, Doubly-Linked List. The recommended patterns to study are: Depth-First Search, Stack.

What are the interview tips for Flatten a Multilevel Doubly Linked List?

Explain your thought process clearly. Consider edge cases like empty lists or lists with no children. Practice writing code on a whiteboard or in a shared document.

What are common mistakes when solving Flatten a Multilevel Doubly Linked List?

Not handling the child pointers correctly. Forgetting to set child pointers to null after flattening.

#430

Flatten a Multilevel Doubly Linked List

Medium

Linked ListDepth-First SearchDoubly-Linked ListDepth-First SearchStack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a depth-first search (DFS) technique to flatten the list in a single pass. This method efficiently handles the child pointers while maintaining the order of nodes.

⚙️

Algorithm

5 steps

1Step 1: Create a stack and push the head node onto it.
2Step 2: While the stack is not empty, pop the top node.
3Step 3: If the node has a next node, push it onto the stack.
4Step 4: If the node has a child, push the child onto the stack and set the child pointer to null.
5Step 5: Link the current node to the previous node.

solution.py26 lines

1# Full working Python code
2class Node:
3    def __init__(self, val=0, prev=None, next=None, child=None):
4        self.val = val
5        self.prev = prev
6        self.next = next
7        self.child = child
8
9class Solution:
10    def flatten(self, head: 'Node') -> 'Node':
11        if not head:
12            return head
13        stack = [head]
14        prev = None
15        while stack:
16            curr = stack.pop()
17            if prev:
18                prev.next = curr
19                curr.prev = prev
20            if curr.next:
21                stack.append(curr.next)
22            if curr.child:
23                stack.append(curr.child)
24                curr.child = None
25            prev = curr
26        return head

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the stack used for DFS, which can store all nodes in the worst case.

1DFS is a powerful technique for tree-like structures.
2Using a stack allows us to manage the nodes efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.