The leaf case and one-child case felt fine, but I fumbled a bit explaining the two-child case live.
Start by explaining the recursive search for the node to delete, then handle the three deletion cases (no child, one child, two children) with the two-child case using the in-order successor (or predecessor). Emphasize maintaining the BST property and discuss time/space complexity, including recursion stack.
Pro tip: Mention that using the in-order successor (leftmost node in right subtree) is a common choice, but the predecessor works equally well; clarify that the choice affects tree shape and could impact future performance. Also, note that deletion in a BST is not guaranteed to keep the tree balanced, so in production you might use a self-balancing tree like a Red-Black or AVL tree.
Restate the problem and confirm assumptions: the tree is a valid BST, the key may or may not exist, and we need to return the new root. Ask if duplicates are allowed or if the tree can be modified in place.
Explain that you will recursively search for the node with the given key, comparing the key with the current node's value to decide whether to go left or right.
Describe the three cases: (1) node with no children: simply remove it; (2) node with one child: replace the node with its child; (3) node with two children: find the in-order successor (or predecessor), copy its value to the node, and recursively delete the successor.
Write clean recursive code, ensuring base cases (null node) are handled. Walk through an example, including edge cases like deleting the root or a leaf.
State that time complexity is O(h) where h is the tree height (O(log n) for balanced, O(n) for skewed). Space complexity is O(h) due to recursion. Discuss iterative alternative to reduce space, and mention self-balancing trees for guaranteed performance.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by defining what a balanced BST and a completely skewed BST are, then compare the time complexities for core operations (search, insert, delete) in each case. Explain how the height of the tree determines the complexity, and conclude with the practical implications for performance.
Pro tip: Mention that balanced BSTs guarantee O(log n) worst-case time, while skewed BSTs degrade to O(n), and note that self-balancing trees like AVL or Red-Black trees are used in practice to avoid this degradation.
Briefly describe a balanced BST (height ~ log n) and a completely skewed BST (height ~ n).
List the main operations: search, insert, and delete, and note that their time complexity depends on tree height.
For balanced BST, operations are O(log n); for skewed BST, they become O(n). Explain why: height difference.
Mention that skewed trees lead to worst-case linear time, which is inefficient for large data, hence self-balancing trees are preferred.
Summarize that balancing maintains efficiency, while skewness can occur with naive insertions (e.g., sorted order) and should be avoided.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, explain the recursive solution's logic and identify the implicit call stack. Then, simulate that stack using an explicit data structure (e.g., stack or queue) and a loop, ensuring you handle state and order of operations correctly. Finally, discuss trade-offs like space complexity and readability.
Pro tip: Mention that while recursion is often cleaner, iterative solutions can avoid stack overflow and sometimes improve performance. Also, note that the choice depends on the problem and constraints.
Clearly state what the recursive function does, its base case, and how it combines results. Identify the state that changes with each call.
Recognize that recursion uses the call stack to store local variables and return addresses. Determine what information needs to be stored to simulate this.
Select a stack (for depth-first) or queue (for breadth-first) to mimic the call stack. Sometimes a simple loop with variables suffices for tail recursion.
Replace recursive calls with push/pop operations. Ensure the order of processing matches the original recursion (e.g., for tree traversals, push right then left for preorder).
Walk through an example to verify correctness. Compare time and space complexity with the recursive version, noting any improvements or trade-offs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the definition of duplicate keys in the BST (e.g., whether they are allowed and how they are stored). Then explain the deletion strategy: either delete all occurrences, delete one occurrence, or handle duplicates by augmenting nodes with counts. Discuss trade-offs and choose an approach based on requirements.
Pro tip: Demonstrate awareness that duplicate handling is a design decision: many BST implementations disallow duplicates, but if allowed, a count field per node is often the most efficient. Mention that Google values clean, scalable solutions and clear communication of assumptions.
Ask whether duplicates are allowed in the BST and how they are represented (e.g., multiple nodes with same key, or a count field). Confirm the expected behavior when deleting a key with duplicates.
Decide between storing duplicates as separate nodes (e.g., always in the right subtree) or augmenting each node with a count. Discuss the implications for search, insert, and delete.
For separate nodes: either delete all occurrences by repeatedly deleting the key, or delete one occurrence and leave others. For count field: decrement the count and remove the node only when count reaches zero.
Compare time complexity (e.g., O(h) for single deletion vs O(kh) for k duplicates), space overhead, and code complexity. Consider edge cases like deleting the last occurrence or when the node has two children.
Based on typical requirements (e.g., efficiency, simplicity), recommend using a count field per node as it handles duplicates elegantly and keeps the tree balanced. If duplicates as separate nodes are required, specify the deletion order.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I knew the high-level answer (rotations, rebalancing after deletion) but I was pretty vague on Red-Black specifics.
Start by explaining the standard BST deletion cases, then highlight that self-balancing trees must restore balance after deletion via rotations and recoloring. Compare AVL and Red-Black tree deletion in terms of rebalancing strategies and complexity, and discuss the trade-offs in performance.
Pro tip: Emphasize that while deletion in self-balancing trees is more complex, it guarantees O(log n) worst-case time, which is crucial for real-time systems. Mention that Red-Black trees often require fewer rotations than AVL trees during deletion, making them preferable for write-heavy workloads.
Briefly outline the three cases in a plain BST: deleting a leaf, a node with one child, and a node with two children (using inorder successor/predecessor).
After deletion, self-balancing trees may violate their balance invariants (height balance for AVL, color properties for Red-Black), so they must perform rotations and/or recoloring to restore balance.
Describe how AVL trees rebalance after deletion by checking balance factors along the path to the root and applying single or double rotations. Note that deletion may require O(log n) rotations.
Explain that Red-Black trees use color flips and rotations to maintain properties. Deletion often involves fixing double-black nodes through cases, with at most O(1) rotations (though O(log n) recoloring).
Summarize key differences: AVL trees are more strictly balanced, leading to faster lookups but potentially more rotations on deletion; Red-Black trees have looser balance, resulting in fewer rotations and better performance for insert/delete-heavy workloads.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.