In-order traversal gives you a sorted sequence, so I just did that and counted up to k.
Use an in-order traversal of the BST to visit nodes in ascending order, keeping a counter to stop at the kth node. This yields O(k) time and O(h) space, which is optimal for a single query. If multiple queries are expected, discuss augmenting nodes with subtree sizes for O(log n) per query.
Pro tip: Mention that in-order traversal can be done iteratively with a stack to avoid recursion overhead and to handle large trees gracefully. Also, clarify the 1-indexed nature of k and handle edge cases like k > number of nodes.
Confirm that k is 1-indexed, the tree is a valid BST, and discuss edge cases such as k being larger than the number of nodes. Ask if multiple queries will be made to decide between simple traversal and augmented tree.
Decide between recursive and iterative in-order traversal. Iterative is often preferred for its explicit stack and early termination, but recursive is simpler if the tree depth is manageable.
Traverse the tree in-order, incrementing a counter at each node. When the counter equals k, return the node's value immediately to avoid unnecessary traversal.
Explain that time complexity is O(k) in the best case (early stop) and O(n) in the worst case, while space is O(h) for the stack, where h is the tree height.
If multiple queries are expected, propose augmenting each node with the size of its left subtree to find the kth smallest in O(log n) time per query.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.