Flatten a Multilevel Doubly Linked List

Data Structures
Medium
Google
77.8K views

Given a doubly linked list that may contain child pointers, flatten the list such that all nodes appear in a single-level, doubly linked list. The child list should appear immediately after the parent node.

Why Interviewers Ask This

Google interviewers ask this to evaluate a candidate's mastery of pointer manipulation and recursive thinking within complex data structures. They specifically test your ability to handle edge cases like nested children while maintaining strict doubly linked list invariants. This problem reveals whether you can manage stack depth, memory references, and traversal logic without corrupting the original structure.

How to Answer This Question

1. Clarify Requirements: Confirm if the input is mutable and how deeply nested the lists might be. 2. Visualize Structure: Draw a small example with two levels of nesting to trace the 'next' and 'prev' pointer changes mentally. 3. Choose Strategy: Decide between recursion (cleaner code) or an iterative stack approach (better for deep nesting). 4. Define Base Cases: Explicitly state what happens when a node has no child or no next pointer. 5. Execute Traversal: Explain that upon finding a child, you must save the current 'next' node, link the child as the new 'next', and update the child's 'prev'. 6. Reconnect: After processing the child subtree, retrieve the saved 'next' node and link it properly to the end of the child list. 7. Validate: Walk through your logic on a multi-level example to ensure no pointers are lost or circular references created.

Key Points to Cover

  • Explicitly mention handling the 'saved next' pointer to prevent data loss during reconnection
  • Demonstrate understanding of O(N) time complexity by visiting each node once
  • Clarify the distinction between recursion stack depth and total node count for space complexity
  • Show step-by-step pointer updates for both 'next' and 'prev' directions
  • Reference Google's focus on writing clean, maintainable code that handles edge cases

Sample Answer

To flatten this multilevel doubly linked list, I would first analyze the structure to understand the depth of nesting. The core challenge is preserving the doubly linked properties—specifically ensuring that both 'next' and 'prev' pointers remain consistent after restructuring. I propose using a recursive depth-first search approach because it naturally handles arbitrary nesting depths. My algorithm starts at the head. If a node has a child, I temporarily store its original 'next' pointer. Then, I set the child as the new 'next' node and update the child's 'prev' pointer to the current node. Next, I recursively call the function on this new child node to flatten the entire sub-list. Once the recursion returns, the current node's 'next' will point to the tail of the flattened child list. At this point, I reconnect the stored original 'next' pointer to the end of the child list and update its 'prev' pointer accordingly. If there is no child, I simply move to the next node. This ensures every node is visited exactly once, resulting in O(N) time complexity where N is the total number of nodes, and O(D) space complexity for the recursion stack where D is the maximum depth. This approach aligns well with Google's emphasis on clean, efficient algorithms that handle edge cases robustly.

Common Mistakes to Avoid

  • Forgetting to update the 'prev' pointer of the original next node when reconnecting it to the child list
  • Creating infinite loops by failing to set the tail of the child list's next pointer to null
  • Using excessive auxiliary space instead of leveraging the existing list structure or recursion stack
  • Ignoring deeply nested scenarios where a child itself contains another child list

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