I know LCA for binary trees pretty well but the N-ary version tripped me up for a minute because you can't just go left/right anymore.
Use a post-order DFS that returns the current node if it matches p or q, otherwise recursively searches children. If two children return non-null, the current node is the LCA; otherwise, propagate the non-null result upward. This single traversal finds the LCA in O(N) time and O(H) space.
Pro tip: Clarify that the algorithm assumes both p and q exist in the tree; if not, you may need a separate check or a modified return value. Mentioning this edge case shows attention to detail and avoids incorrect assumptions.
Write a function that takes a node and returns the LCA if found, or the node itself if it matches p or q, or null otherwise.
If the current node is null, or equals p or q, return the current node immediately.
For each child, call the function and collect the results. Count how many non-null results are returned.
If two or more children return non-null, the current node is the LCA. If exactly one child returns non-null, return that result. If none, return null.
Explain that each node is visited once, giving O(N) time, and recursion depth is O(H) where H is tree height, so O(H) space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.