Flatten a Multilevel Doubly Linked List (Stack)

Data Structures
Medium
IBM
85.1K views

Flatten a doubly linked list that may contain child pointers. Use a Stack to manage the next level of lists to be processed.

Why Interviewers Ask This

IBM evaluates this question to assess a candidate's mastery of pointer manipulation and recursive thinking in complex data structures. It specifically tests the ability to manage state transitions using an explicit stack, ensuring candidates can handle non-linear linked list topologies without causing memory leaks or infinite loops.

How to Answer This Question

1. Clarify the input: Confirm if the head node is null and define the exact structure of the child pointer. 2. Propose the iterative stack strategy: Explain that you will traverse the main list, pushing 'next' nodes onto a stack whenever a 'child' is encountered, then connecting the current node to its child. 3. Detail the re-linking logic: Describe how to pop from the stack to resume traversal after the child subtree is flattened, ensuring all pointers (prev, next, child) are correctly updated. 4. Discuss edge cases: Mention handling empty lists, nodes with only children, or nested levels deeper than two. 5. Analyze complexity: Explicitly state O(N) time complexity since every node is visited once, and O(N) space for the worst-case stack depth.

Key Points to Cover

  • Explicitly mention using a stack to defer processing of 'next' nodes
  • Correctly update both 'next' and 'prev' pointers during re-linking
  • Ensure the 'child' pointer is set to null after processing
  • Analyze time complexity as O(N) and space as O(N)
  • Handle the edge case where the input list is empty

Sample Answer

I would approach flattening this multilevel doubly linked list using an iterative method with an explicit stack to ensure clarity and avoid recursion depth issues. First, I initialize a stack and start traversing from the head. As I iterate through each node, if I encounter a node with a child pointer, I save the current 'next' node by pushing it onto the stack, then redirect the current node's 'next' to point to its 'child'. Crucially, I must update the 'prev' pointer of the new next node to point back to the current node and clear the child pointer to zero. Once I reach the end of the current path, if the stack is not empty, I pop the saved 'next' node and continue traversal. This ensures that any deferred branches are processed immediately after their parent branch is fully flattened. For example, if we have A -> B(child C) -> D, when at B, I push D, link B to C, and later resume with D. This approach guarantees O(N) time complexity as each node is visited exactly once, and O(N) space for the stack in the worst case where all nodes form a single vertical chain. This solution aligns well with IBM's focus on robust, memory-efficient algorithms for enterprise-grade systems.

Common Mistakes to Avoid

  • Forgetting to update the 'prev' pointer of the newly connected child node
  • Leaving the 'child' pointer active, which violates the problem constraints
  • Using recursion without considering stack overflow risks on deep trees
  • Failing to check if the stack is empty before popping, causing runtime errors

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