I went with the recursive in-order approach first, threading a 'prev' pointer through the traversal.
Use an in-order traversal to visit nodes in sorted order, relinking each node's left pointer to the previous node and right pointer to the next. Maintain a prev pointer and a head pointer, updating them as you traverse. For duplicates, decide on a consistent policy (e.g., keep them in the list) and explain how the BST property handles them (e.g., duplicates go to the right subtree).
Pro tip: Clarify upfront whether duplicates are allowed and how they are stored in the BST; this shows you think about edge cases and data integrity. Also, mention that the conversion is in-place with O(n) time and O(h) space for recursion, and discuss iterative Morris traversal for O(1) space if asked.
Ask if duplicates are allowed, how they are handled in the BST (e.g., right subtree), and confirm in-place conversion with no new nodes. Also confirm return type (head of list).
Use in-order traversal (recursive or iterative) to process nodes in sorted order. Maintain a 'prev' pointer to the last processed node and a 'head' pointer for the first node.
For each node, set node.left = prev (if prev exists, set prev.right = node). Update prev to current node. At the end, set prev.right = null to terminate the list.
Explain that duplicates are included in the list in sorted order. If the BST stores duplicates in the right subtree, in-order traversal will visit them after the original, preserving stability.
State time O(n) and space O(h) for recursion (or O(1) with Morris). Discuss edge cases: empty tree, single node, skewed tree, and duplicates.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Not too bad once the base solution is done.
First, clarify the original problem and the existing linear doubly linked list implementation. Then, explain how to adjust the pointers so that the last node's next points to the head and the head's prev points to the last node, ensuring all operations maintain circularity. Finally, discuss edge cases and how to test the modified structure.
Pro tip: Emphasize that circularity simplifies edge cases like insertion at the beginning or end, but requires careful handling to avoid infinite loops during traversal. Mention that you would add a sentinel node or a size counter to make operations cleaner.
Restate the original problem and the existing linear doubly linked list solution to ensure alignment with the interviewer.
Determine which pointers need modification: the last node's next should point to the head, and the head's prev should point to the last node.
Modify insertion, deletion, and traversal methods to maintain the circular structure, handling cases like empty list and single node.
Discuss how to avoid infinite loops during traversal and ensure operations work correctly when the list is empty or has one node.
Outline a testing strategy: unit tests for each operation, checking forward and backward traversal, and verifying circular links.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying that O(1) auxiliary space means no recursion stack or explicit stack, and that Morris traversal achieves this by temporarily modifying the tree using threaded links. Then walk through the algorithm step-by-step, emphasizing how it restores the tree and handles edge cases.
Pro tip: Mention that while Morris traversal is space-efficient, it temporarily mutates the tree, which can be problematic in concurrent or read-only scenarios—showing awareness of trade-offs beyond just complexity.
Confirm that 'O(1) auxiliary space' excludes recursion and explicit stacks, and that the output list/array is not counted. State that Morris traversal achieves this by using the tree's null right pointers as temporary threads.
Describe how for each node, you find its inorder predecessor and set the predecessor's right pointer to the current node, creating a temporary link. This allows returning to the current node after traversing the left subtree.
Outline the loop: while current is not null, if no left child, visit current and move right; else find predecessor, if predecessor's right is null, set it to current and move left; if it points to current, reset it to null, visit current, and move right.
Emphasize that the temporary links are removed when revisiting the predecessor, restoring the tree. Mention handling of single-node trees, skewed trees, and the fact that each edge is traversed at most twice, giving O(n) time.
Acknowledge that Morris traversal modifies the tree temporarily, which may not be suitable for concurrent access. Compare with recursive/stack-based approaches that use O(h) space but are simpler and non-mutating.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.