My first instinct was BFS and that was a mistake.
Clarify that the tree is weighted and the path length is the sum of edge weights. Then explain that the longest path (diameter) can be found using two DFS/BFS passes: first from any node to find the farthest node A, then from A to find the farthest node B; the distance between A and B is the diameter. Alternatively, use tree DP to compute the maximum downward path and combine the two longest downward paths at each node.
Pro tip: Mention that the two-pass method works because the farthest node from any node is an endpoint of a diameter, and that tree DP is more general for other problems. Also, discuss handling negative weights: if edges can be negative, the two-pass method fails, and you need tree DP with careful initialization.
Confirm that the tree is weighted, edges are undirected, and path length is the sum of edge weights. Ask if edge weights can be negative.
Select either the two-pass DFS/BFS method (for non-negative weights) or tree DP (for general weights). Explain the reasoning behind your choice.
For two-pass: pick any node, run DFS to find farthest node A, then from A run DFS to find farthest node B; return distance A-B. For tree DP: for each node, compute the longest downward path and the sum of the two longest downward paths from its children; track the maximum.
State that both approaches run in O(n) time and O(n) space, where n is the number of nodes.
Mention handling of a single node (diameter 0), negative weights (if applicable), and large trees (recursion depth).
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