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The promotion rule is what makes this feel different from a standard delete question.
Clarify the deletion semantics and root-deletion behavior, then propose a post-order traversal that computes the height of each subtree after deletions, handling promoted children and forest cases. Discuss time and space complexity, aiming for O(n) time and O(h) space.
Pro tip: Mention that deletions can be processed in a single post-order traversal without physically modifying the tree, by treating deleted nodes as transparent and promoting their children. This shows you can optimize both time and space.
Ask questions to confirm: Are node IDs unique? What does 'preserving left/right positions' mean exactly? If a deleted node has two children, do they both get promoted, and how are they ordered? What if the root is deleted—does each child become a root of a new tree?
Design a post-order traversal that returns the height of the subtree rooted at the current node after deletions. If the current node is deleted, combine the heights of its children appropriately (e.g., if both children exist, the promoted subtree's height is max(leftHeight, rightHeight) + 1? Actually, need to think: when a node is deleted, its children are promoted to its parent. So the height contribution from that subtree is the maximum of the heights of its children, but if both children exist, they become siblings under the parent, so the height is max(leftHeight, rightHeight) + 1? Wait, careful: The height of a tree is the number of nodes on the longest path from root to leaf. If a node is deleted, its children take its place. So if the node had two children, they both become children of the parent, so the height of the subtree rooted at the parent will be 1 + max(height of left child's subtree, height of right child's subtree) but note that the children are now directly attached to the parent, so the edge from parent to child counts. So if the deleted node had children, the height contributed is max(height(left), height(right)) + 1? Actually, if the deleted node is at depth d from the parent, and it had children, those children are now at depth d+1 from the parent? Let's think: parent -> deleted node -> child. After deletion, parent -> child. So the child moves up one level. So the height of the subtree rooted at the parent is 1 + max(height of child subtrees). But if the deleted node had no children, it contributes 0. So the recursive function should return the height of the subtree rooted at the current node after deletions, considering that if the current node is deleted, we need to merge its children's heights. However, if the current node is deleted, we cannot simply return max(left, right) because the children become siblings under the parent, so the height from the parent's perspective is 1 + max(left, right) if at least one child exists, else 0. But the function returns the height of the subtree rooted at the current node. If the current node is deleted, the subtree rooted at the current node is effectively replaced by its children, so the height of that 'subtree' (now a forest of up to two trees) is max(left, right). But when attached to the parent, the parent will add 1. So the function should return the height of the forest resulting from the current node's subtree if the current node is deleted. That is, if deleted, return max(height(left), height(right)) (or 0 if no children). If not deleted, return 1 + max(height(left), height(right)). This works because if not deleted, the node itself adds 1 to the height. If deleted, the node is removed, so the height is just the max of the children's heights. But careful: if the node is deleted and has two children, they both become children of the parent, so the height from the parent's perspective is 1 + max(left, right). But the function returns the height of the subtree rooted at the current node after deletion, which is max(left, right) because the current node is gone. Then the parent will add 1 if the parent is not deleted. So this is consistent.
If the root is deleted, the result is a forest of up to two trees (the root's children). The overall height is the maximum height among these trees. So after processing the root, if it is deleted, return max(height(left), height(right)) as the final answer; otherwise return 1 + max(height(left), height(right)).
Time complexity is O(n) because each node is visited once. Space complexity is O(h) for the recursion stack, where h is the height of the original tree (worst case O(n) for skewed tree).
Consider empty tree, deleting all nodes, deleting a leaf, deleting a node with one child, and deleting the root. Also consider if the tree is a single node and it is deleted—resulting forest is empty, so height 0.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the child-promotion deletion rule and confirm assumptions about the tree structure and deletion semantics. Then, design a recursive search that explores all valid deletion sets, using pruning based on current height and remaining deletions, and deduplicate results by canonicalizing sets. Finally, analyze correctness and complexity, discussing trade-offs between exhaustive search and optimizations.
Pro tip: Explicitly state your assumptions about the deletion rule and tree properties upfront; this shows you can handle ambiguity and sets the stage for a focused solution. Also, mention that you would test with small trees to validate the rule before scaling up.
Restate the child-promotion deletion rule, confirm tree properties (e.g., node IDs unique, binary tree), and define what 'height' means (e.g., number of edges on longest root-to-leaf path). Ask clarifying questions if needed.
Use DFS to explore deleting or keeping each node, tracking current height and deletions used. Prune branches where height already exceeds k or deletions exceed n, and where remaining nodes cannot achieve height k.
Store each valid deletion set in a hash set after sorting the node IDs to avoid duplicates. At the end, convert the set to a list of sorted lists.
Explain that the search exhaustively considers all subsets of nodes up to size n, and pruning only removes branches that cannot lead to a valid solution. Thus, every valid set is found exactly once.
State worst-case time complexity O(2^m * m) where m is number of nodes, but pruning reduces practical runtime. Space complexity O(2^m) for storing results. Mention potential optimizations like memoization on subtree states.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.