Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026Remote

Summary

Meta SWE coding round, one problem the whole time. Classic tree manipulation question but the circular linking part tripped me up more than I expected.

Questions Asked (1)

Given the root of a binary search tree, convert it in place to a sorted circular doubly linked list. Left pointers serve as predecessors, right pointers as successors, and the list wraps around so the smallest and largest elements point to each other. Return the pointer to the smallest element.

Algorithms & Data Structures

Author's notes

I knew in-order traversal was the right move pretty fast.

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AI HintsAI Generated

Suggested Approach

Use an in-order traversal to visit nodes in sorted order, relinking left and right pointers as you go to form the doubly linked list. After traversal, connect the head (smallest) and tail (largest) to make it circular, then return the head.

Pro tip: Clarify whether the conversion should be done in-place and whether recursion is acceptable; if not, mention that an iterative in-order traversal with a stack can achieve O(1) extra space (excluding stack).

1. Clarify requirements and constraints

Confirm that the transformation is in-place, that left/right pointers become predecessor/successor, and that the list must be circular. Ask about recursion depth limits or space constraints.

2. Plan in-order traversal with relinking

Perform an in-order traversal (recursive or iterative) to process nodes in ascending order. Maintain a 'prev' pointer to the last processed node and a 'head' pointer for the smallest node.

3. Relink pointers during traversal

For each node, set its left pointer to 'prev' and, if 'prev' exists, set 'prev.right' to the current node. Update 'prev' to the current node.

4. Make the list circular

After traversal, connect the head's left pointer to the tail (last processed node) and the tail's right pointer to the head.

5. Return the head and test edge cases

Return the head (smallest element). Test with empty tree, single node, and skewed trees to ensure correctness.

Key Points to Mention

In-order traversal yields sorted order for a BST.
Maintain prev and head pointers to relink nodes efficiently.
Handle edge cases: empty tree, single node, and skewed trees.
Time complexity: O(n) since each node is visited once.
Space complexity: O(h) for recursion stack (or O(1) if iterative with parent pointers).
Ensure the list is circular by connecting head and tail.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.