The base case felt fine, recursive DFS, return the node when you find either target, bubble up.
Clarify whether the tree is a binary search tree or a general binary tree, and whether nodes have parent pointers. For a general binary tree, use a recursive post-order traversal that returns the node if it matches p or q, otherwise recurses into left and right subtrees; if both return non-null, the current node is the LCA. Discuss time and space complexity, and consider iterative or parent-pointer alternatives if applicable.
Pro tip: Always ask clarifying questions about the tree type and node structure before coding; this shows you think about edge cases and can lead to a more optimal solution. Also, mention that the recursive solution uses O(h) space due to the call stack, and discuss how to handle it iteratively if needed.
Ask whether the tree is a binary search tree or a general binary tree, and whether nodes have parent pointers. Confirm that p and q are guaranteed to be in the tree.
For a general binary tree, use a recursive post-order traversal. For a BST, use the BST property to traverse from the root. If parent pointers exist, find the intersection of paths to the root.
Write clean code for the chosen approach. For recursion, define a helper that returns the LCA or null. For BST, iteratively move left or right based on node values.
State the time complexity (O(n) for general tree, O(h) for BST) and space complexity (O(h) for recursion, O(1) for iterative BST).
Walk through edge cases: one node is ancestor of the other, nodes in different subtrees, skewed tree, etc. Verify the solution handles them.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by contrasting the general binary tree LCA approach (which requires traversing both subtrees) with the BST property that allows directional decision-making. Then explain how to leverage the BST ordering to find the split point where both nodes diverge, achieving O(h) time and O(1) space. Finally, discuss edge cases and trade-offs compared to the general solution.
Pro tip: Emphasize that the BST optimization reduces both time and space complexity by eliminating the need for recursion or parent pointers, and mention that this is a common follow-up to the general LCA problem at companies like Meta.
Briefly explain that LCA in a general binary tree typically uses recursion or parent pointers, visiting nodes until both targets are found. This sets the baseline for optimization.
Explain that in a BST, for any node, all values in the left subtree are smaller and all in the right are larger. This allows us to decide which subtree to explore based on the values of the two target nodes.
Start at the root and compare its value with the two targets. If both targets are smaller, move left; if both are larger, move right; otherwise, the current node is the LCA (the split point).
State that time complexity is O(h) where h is the height (O(log n) for balanced BST, O(n) worst-case), and space is O(1) iterative. Mention edge cases: one node is ancestor of the other, duplicate values, or empty tree.
Highlight that the BST approach is more efficient in both time and space, but relies on the BST invariant. If the tree is not a BST, the general approach is necessary.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Explain that with parent pointers, you can find the LCA by first computing the depths of both nodes, then aligning them by moving the deeper node up, and finally moving both nodes up in tandem until they meet. This avoids traversing from the root and achieves O(h) time and O(1) space.
Pro tip: Mention that this approach is essentially the two-pointer technique for linked lists, and highlight that it's optimal because it uses constant extra space and doesn't require root access.
Traverse from each node up to the root using parent pointers to determine the depth of each node.
Move the deeper node up by the difference in depths so that both nodes are at the same level.
Move both nodes up one step at a time until they point to the same node, which is the LCA.
Consider cases where one node is an ancestor of the other, or when nodes are the same; the algorithm naturally handles these.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Acknowledge that the in-memory LCA solution doesn't scale, then propose an external-memory approach such as storing parent pointers on disk and using binary lifting with disk-based jumps, or leveraging Euler tour + RMQ with external memory data structures. Discuss trade-offs between preprocessing time, query time, and I/O cost, and mention distributed alternatives like MapReduce or Spark if the tree is massive.
Pro tip: Emphasize that the key challenge is random access to parent pointers; propose a B-tree or LSM-tree based storage to minimize disk seeks, and quantify the I/O complexity (e.g., O(log N) disk reads per query). This shows you think about system-level constraints, not just algorithms.
Ask about tree size, available memory, disk space, query frequency, and whether the tree is static or dynamic. This determines whether you need a one-time preprocessing or an online solution.
Explain that the in-memory LCA algorithms (e.g., binary lifting, Euler tour + RMQ) require O(N) memory and random access, which is infeasible when N exceeds memory. The main bottleneck is random disk I/O.
Suggest storing the tree in a disk-based structure (e.g., B-tree) and adapting binary lifting: store jump pointers on disk and perform O(log N) disk reads per query. Alternatively, use a disk-based Euler tour with a sparse table stored in blocks.
Compare preprocessing time, query time, and I/O cost. Mention caching frequently accessed nodes, using compression, or partitioning the tree if it's too large for a single machine.
If the tree is extremely large (e.g., web graph), propose a distributed approach like MapReduce to compute LCA in parallel, or use a graph processing framework (e.g., Pregel) to precompute ancestor information.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.