The classic mistake I almost made: I started writing the DFS to return the diameter directly.
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, and the diameter is the maximum of all such paths.
Pro tip: Clarify whether the diameter is measured in edges or nodes, and mention that the path may or may not pass through the root. Also, discuss handling edge cases like an empty tree or a single node.
Confirm that the diameter is the number of edges on the longest path between any two nodes, and that the path may not pass through the root.
Design a function that returns the height of a subtree (max edges from root to leaf) and updates a global variable for the maximum diameter.
At each node, calculate the sum of the heights of its left and right subtrees. This represents the longest path passing through that node.
Compare the current node's diameter with the global maximum and update if larger.
Return 1 + max(left height, right height) to the parent, and finally return the global maximum diameter.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Meta apparently loves this as a follow-up and yeah, they asked it.
Generalize the binary tree diameter algorithm by computing, for each node, the two largest depths among its children's subtrees. The diameter is the maximum of (sum of top two depths + 1) over all nodes, and the function returns the maximum depth for parent computations.
Pro tip: Emphasize that the core idea remains the same—tracking the two longest downward paths—but the implementation must handle an arbitrary number of children, so using a list or heap to find the top two depths is key. Also, mention that the diameter may or may not pass through the root, so you must consider all nodes.
Confirm that the diameter is the number of edges on the longest path between any two nodes, and that the tree is N-ary with no parent pointers. Ask about input size to discuss recursion depth and potential iterative solutions.
Design a function that returns the maximum depth (in edges) from the current node down to a leaf. At each node, compute the depths of all children's subtrees.
For each node, find the two largest depths among its children. The longest path through this node has length (largest + second largest + 2) if there are at least two children, or (largest + 1) if only one child.
Maintain a global variable for the maximum diameter seen so far. At each node, update it with the path length computed from the top two depths.
Return 1 + the largest child depth (or 0 if leaf) to the parent, enabling the parent to compute its own top two depths.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First clarify the definition of 'strictly monotonic' (increasing or decreasing) and whether the path must be a simple downward path (from ancestor to descendant) or can go up and down. Then propose a DFS that, for each node, computes the longest increasing and decreasing paths starting at that node, combining them appropriately if paths can bend at a node. Analyze time and space complexity, and discuss trade-offs between recursive and iterative implementations.
Pro tip: Mention that in a BST, an in-order traversal yields sorted values, so a strictly increasing path corresponds to a sequence of nodes visited in in-order—this insight can simplify the problem and impress the interviewer.
Ask whether the path must be strictly increasing, strictly decreasing, or either; whether it must be a simple downward path (ancestor to descendant) or can change direction; and whether the path can skip levels.
For each node, define dp_inc[node] = longest strictly increasing path starting at node going downward, and dp_dec[node] = longest strictly decreasing path starting at node going downward. Recurrence: dp_inc[node] = 1 + max(dp_inc[child] for child with value > node.value), similarly for dp_dec.
If paths can change direction at a node, the longest monotonic path through a node is dp_inc[node] + dp_dec[node] - 1 (if both directions are allowed) or max(dp_inc[node], dp_dec[node]) if only one direction. Track the global maximum during DFS.
Write a recursive DFS that returns both dp_inc and dp_dec for each node, updating a global max. Analyze time complexity O(n) and space O(h) for recursion stack, where h is tree height.
Consider iterative implementation to avoid recursion depth issues, handle empty tree, single node, and duplicate values (strictness). Discuss whether the BST property can be leveraged for optimization.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.