The two-BFS trick is something I'd seen before but blanked on the second pass.
Use two BFS/DFS traversals: first from any node to find one endpoint of the diameter, then from that endpoint to find the other endpoint and reconstruct the path using parent pointers. This approach runs in O(n) time and O(n) space, which is optimal for a tree.
Pro tip: Mention that the two-pass BFS/DFS method works because trees are acyclic and connected, and that you can reconstruct the path by storing parents during the second traversal. Also, note that if the tree is very large, an iterative BFS avoids recursion depth issues.
Confirm that the tree is undirected, connected, and acyclic, and that nodes are uniquely identifiable. Ask if the tree is given as an adjacency list or edges, and whether recursion depth is a concern.
Run BFS or DFS from any arbitrary node (e.g., node 0) to find the farthest node A. This node is guaranteed to be an endpoint of a diameter.
Run BFS or DFS from node A, tracking parent pointers, to find the farthest node B. The path from A to B is the diameter; reconstruct it by following parents from B back to A.
State that time complexity is O(n) and space is O(n). Discuss edge cases: single node (diameter length 0), two nodes, and a star graph where the diameter is between two leaves.
Walk through a small example (e.g., a tree with 5 nodes) to verify the algorithm and path reconstruction. Mention that you would test with a line graph, balanced tree, and star graph.
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