I knew the diameter problem well enough but the tuple-return constraint threw me a bit.
Use a post-order DFS that returns a tuple (height, diameter) for each subtree. At each node, compute the height as 1 + max(left_height, right_height) and the diameter as the maximum of left_diameter, right_diameter, and left_height + right_height. Return the tuple upward, ensuring no global state is used.
Pro tip: Clarify that the diameter is measured in edges, not nodes, and that the path may or may not pass through the root. Mention that the tuple approach elegantly avoids global variables and is thread-safe.
Create a function that takes a node and returns a tuple (height, diameter). For a null node, return (0, 0) or (-1, 0) depending on edge-count convention.
Recursively call the function on the left and right children to obtain their (height, diameter) tuples.
Calculate the current height as 1 + max(left_height, right_height). Compute the diameter through the current node as left_height + right_height (if height is in edges) or left_height + right_height + 2 (if height is in nodes).
The best diameter for the current subtree is the maximum of the left diameter, right diameter, and the diameter through the current node. Return (current_height, best_diameter).
After the initial call on the root, the diameter is the second element of the returned tuple. Return that as the final result.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by contrasting the recursive call stack (which only holds the current path, O(h)) with an explicit list that accumulates every visited node, leading to O(n) space. Then explain that the worst-case space inflation occurs for skewed trees (e.g., a chain), where h = n, and for balanced trees the list still grows to n, but the relative overhead is larger. Finally, discuss the trade-off: explicit lists simplify iteration but sacrifice memory efficiency, and suggest alternatives like parent pointers or iterative DFS with a stack.
Pro tip: Mention that in an interview, you should not just state the complexity but also quantify the constant factors: a list of n nodes stores n references, while the call stack stores only h frames, each with local variables. This shows you understand memory layout, not just Big-O.
Explain that recursive DFS uses the call stack, which holds at most one frame per level of the current path, so space is O(h) where h is tree height.
Describe how storing traversal state in a list means appending every visited node (or state) to a list, so the list grows to the total number of nodes, n, giving O(n) space.
For a skewed tree (h = n), both methods are O(n), but for a balanced tree (h = log n), the list is O(n) while recursion is O(log n), making the inflation most costly in balanced or bushy trees.
Acknowledge that explicit lists can simplify code (e.g., for backtracking or path reconstruction) but at a memory cost; suggest alternatives like parent pointers, iterative DFS with an explicit stack (which is still O(h) if done correctly), or Morris traversal for O(1) space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the diameter of a binary tree is the longest path between any two nodes, typically computed via postorder traversal. Then, explain how to simulate the recursive postorder traversal iteratively using an explicit stack while maintaining O(h) auxiliary space by storing only the path from root to current node. Finally, describe how to compute and update the diameter during this traversal.
Pro tip: Emphasize that the explicit stack should store nodes along the current path, not all visited nodes, to achieve O(h) space. Also, mention that this approach avoids recursion depth limits and is more robust for deep trees.
Explain that the diameter is the maximum of left height + right height at each node. Describe the standard recursive postorder solution that returns height and updates a global diameter.
Use a stack to mimic the call stack, pushing nodes and tracking visited state (e.g., with a last visited pointer or a state flag) to ensure children are processed before the parent.
Ensure the stack only contains nodes along the current root-to-leaf path, not all nodes. Avoid extra data structures that grow with n, such as a full traversal list.
When processing a node, retrieve the heights of its left and right subtrees from the stack or temporary variables, compute the diameter candidate, and update the global maximum.
Discuss empty tree, single node, skewed trees, and confirm time O(n) and space O(h). Mention that h can be n in worst case, but that's the tree height.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the diameter of a binary tree is the longest path between any two nodes, then compare the recursive and iterative solutions. For each, derive time and space complexity by analyzing the number of node visits and the maximum stack/queue size, and justify why they are O(n) time and O(h) or O(n) space.
Pro tip: Emphasize that the iterative solution's space complexity depends on the tree's height if using a stack for post-order traversal, but can be O(n) in the worst case; showing awareness of this nuance demonstrates depth.
State that the diameter is the longest path between any two nodes, and that both recursive and iterative solutions typically compute the height of each subtree while tracking the maximum diameter.
Explain that the recursive DFS visits each node once, giving O(n) time, and uses call stack space proportional to tree height, O(h), which is O(n) in the worst case.
Describe an iterative post-order traversal using a stack, which also visits each node once (O(n) time) and uses O(h) space for the stack, but may require additional data structures for storing heights, potentially O(n) space.
Highlight that both have O(n) time, but space differs: recursive uses call stack, iterative uses explicit stack; both are O(h) in balanced trees and O(n) in skewed trees.
Mention that iterative avoids recursion depth limits but may have higher constant factors; recursive is simpler but risks stack overflow for deep trees.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.