BFS with a queue is the obvious move and I went there immediately, but the tricky part is structuring it so next() only processes one level per call rather than the whole tree upfront.
Use a queue to perform level-order traversal, but process one full level at a time by tracking the number of nodes at the current level. For the iterator, maintain the queue and a count of nodes remaining in the current level; next() returns all values of the current level and prepares the next level, while hasNext() checks if the queue is non-empty. This achieves O(1) amortized time per node and O(width) space, which is optimal for level-order traversal.
Pro tip: Clarify the expected time complexity for next(): it should be O(1) per node on average, but if next() returns all nodes of a level, its time is proportional to the level size. Also, discuss space optimization: the queue holds at most the maximum width of the tree, which is optimal for BFS.
Confirm that next() returns a list of all node values at the current level, and hasNext() indicates if there are more levels. Ask about edge cases like empty tree and whether the tree can be modified during iteration.
Select a queue (FIFO) to perform level-order traversal. Explain that by tracking the number of nodes in the current level, you can process one level at a time without mixing levels.
Maintain a queue of nodes and an integer levelSize representing the number of nodes in the current level. In next(), dequeue levelSize nodes, collect their values, enqueue their children, and update levelSize to the queue size for the next level. hasNext() simply checks if the queue is non-empty.
State that each node is enqueued and dequeued once, so total time is O(N) for N nodes, and next() takes O(levelSize) time. Space is O(W) where W is the maximum width of the tree, which is optimal for BFS. Discuss alternatives like DFS with level tracking, but note BFS is more natural for level-order.
Consider empty tree: hasNext() returns false initially. Also, ensure that after the last level, hasNext() returns false. Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the iterator design: are they independent, nested, or concurrent? Then discuss how to maintain correctness (e.g., snapshotting, versioning, or immutable nodes) and performance (e.g., lazy evaluation, caching, or structural sharing). Finally, analyze trade-offs and propose a solution that balances both, such as using a persistent data structure with path copying.
Pro tip: Mention that Google often values scalability and concurrency; highlight how your solution handles concurrent modifications without locks, e.g., by leveraging immutability or MVCC. Also, briefly discuss how you would test correctness and measure performance.
Ask whether iterators are independent, nested, or concurrent, and whether the tree can be modified during iteration. This determines the necessary guarantees.
Discuss issues like iterator invalidation, consistency, and isolation. For example, if the tree is modified, how do you ensure each iterator sees a consistent view?
Suggest techniques such as immutable nodes, versioning, snapshotting, or copy-on-write. Explain how each ensures that iterators remain valid and see a consistent state.
Discuss how to minimize overhead: lazy evaluation, caching, structural sharing, and avoiding deep copies. Analyze time and space complexity for multiple iterators.
Compare approaches (e.g., locking vs. lock-free, eager vs. lazy) and justify a choice based on the scenario, emphasizing scalability and concurrency.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.