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Pretty standard DFS once you frame it right.
Use a post-order DFS that returns the height of the valid subtree rooted at each node, treating deleted nodes as contributing height 0 and pruning their descendants. At each node, compute 1 + max(height of valid children), but if the node itself is deleted, return 0 immediately. The answer is the height returned for the root, or 0 if the root is deleted.
Pro tip: Clarify the definition of 'deleted' and whether the tree is given as an adjacency list or parent array; then explicitly state that you prune at deleted nodes to avoid traversing disconnected subtrees, which keeps the solution O(n).
Confirm that a deleted node removes its entire subtree, height counts nodes on the longest root-to-leaf path, and the root deletion returns 0. Ask how the tree is represented (e.g., adjacency list, parent array) and whether deleted nodes are marked in a boolean array.
Select a post-order DFS because a node's height depends on the heights of its valid children. This naturally handles pruning: when a node is deleted, return 0 without recursing into its children.
Write a function height(node) that returns 0 if node is deleted; otherwise, compute 1 + max(height(child) for each child). If a child is deleted, its height is 0 and it contributes nothing to the path.
If the root is deleted, return 0 immediately. For a leaf node that is not deleted, height is 1. Ensure that deleted nodes do not propagate their descendants' heights upward.
State that the algorithm visits each node at most once, so time is O(n) and space is O(h) for recursion (or O(n) worst case). Walk through a small example with a deleted internal node to verify pruning.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as finding a set of nodes to delete so that the remaining tree has height ≤ K, minimizing deletions. Use dynamic programming on the tree: for each node, compute the minimum deletions in its subtree for each possible height, then combine children by keeping at most K levels. The answer is the minimum deletions for the root with height ≤ K.
Pro tip: Clarify that 'effective tree' means the tree after deletions, and note that deleting a node also removes its entire subtree. This shows you understand the problem's constraints and avoids misinterpretation.
Confirm that deleting a node removes its entire subtree, and that the height is measured in edges or nodes. Ask if K is given as a non-negative integer.
For each node u, let dp[u][h] be the minimum number of deletions in the subtree rooted at u such that the resulting subtree has height at most h. h ranges from 0 to K.
For a leaf, dp[u][h] = 0 for all h ≥ 0. For an internal node, either delete u (cost = size of subtree) or keep u and combine children: for each child v, we need dp[v][h-1] if we keep the edge, but we can also delete the child (cost = size of v's subtree). The recurrence is dp[u][h] = min( size(u), sum over children of min( dp[v][h-1], size(v) ) ).
Perform a post-order traversal to compute subtree sizes and dp values for all nodes and heights from 0 to K. Use memoization or iterative DP.
The minimum number of additional deletions needed is dp[root][K]. If the original height is already ≤ K, the answer is 0.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.