The tricky part is that the longest path doesn't have to go through the root, which is easy to forget under pressure.
Use a post-order DFS that returns the height of each subtree while updating a global maximum diameter. At each node, the longest path through it is the sum of the heights of its left and right subtrees; the overall answer is the maximum of this value across all nodes.
Pro tip: Clarify that the diameter is measured in edges, not nodes, and mention that the path may or may not pass through the root. Also, note that the tree could be skewed, so recursion depth might be O(n) and an iterative approach may be needed to avoid stack overflow.
Confirm that the diameter is the number of edges on the longest path between any two nodes, and that the path does not necessarily pass through the root. Ask about constraints (e.g., tree size) to discuss recursion limits.
Define a function that returns the height of a subtree (max edges from the node to a leaf). At each node, compute the left and right heights.
At each node, the candidate diameter is left_height + right_height. Update a global maximum with this value.
Return 1 + max(left_height, right_height) to the parent, representing the height of the current subtree.
State that the time complexity is O(n) since each node is visited once, and space complexity is O(h) for recursion stack, where h is the tree height (O(n) worst case).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.