I went straight for in-order traversal for the smallest, which is the obvious move.
Use an in-order traversal to get the k-th smallest and a reverse in-order traversal to get the k-th largest, both in O(h + k) time. Alternatively, augment the BST with subtree sizes to achieve O(h) time for each query. Discuss trade-offs between time complexity and extra space.
Pro tip: Mention that if the BST is frequently modified, augmenting nodes with subtree sizes allows O(log n) queries, which is ideal for dynamic scenarios. Also, clarify that k is 1-indexed and handle edge cases like k > n.
Confirm whether the BST is static or dynamic, and whether k is 1-indexed. Ask about constraints on n and k, and if multiple queries are expected.
For a single query, use in-order traversal for k-th smallest and reverse in-order for k-th largest. For multiple queries, consider augmenting the BST with subtree sizes.
Write a recursive or iterative in-order traversal that stops at the k-th element. For k-th largest, traverse right subtree first.
Time: O(h + k) for traversal, O(h) for augmented BST. Space: O(h) for recursion stack, O(n) extra for augmentation.
Check if k is valid (1 ≤ k ≤ n). If not, return null or throw an exception. Also consider duplicate values if allowed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.