Got through this fine with the standard recursive post-order approach.
Start by clarifying the problem and constraints, then propose a recursive DFS solution that traverses the tree and returns the LCA when both nodes are found in different subtrees or when the current node is one of the targets. Discuss time and space complexity, and consider edge cases like skewed trees.
Pro tip: Mention that the recursive solution uses O(h) space due to the call stack, and if the tree is very deep, an iterative approach with parent pointers can avoid stack overflow. Also, note that the problem guarantees both nodes exist, so you don't need to handle missing nodes.
Confirm assumptions: nodes p and q are guaranteed to exist, tree is not necessarily BST, and we need the deepest common ancestor. Ask if the tree can be modified or if we can use extra space.
Explain that you'll do a post-order traversal: recursively search left and right subtrees. If both return non-null, current node is LCA; if one returns non-null, propagate it upward.
Trace the algorithm on a small tree to demonstrate correctness, showing how the LCA is identified when p and q are in different subtrees or when one is an ancestor of the other.
State that time complexity is O(n) since each node is visited once, and space complexity is O(h) for the recursion stack, where h is tree height.
Mention edge cases: p or q is the root, tree is skewed, or p is ancestor of q. Briefly mention iterative solutions using parent pointers or path finding if recursion depth is a concern.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Explain that in a BST, the LCA of two nodes is the first node whose value lies between the two target values (inclusive). Then describe a top-down traversal from the root, moving left or right based on comparisons, achieving O(h) time and O(1) space.
Pro tip: Mention that this approach avoids the need for parent pointers or storing paths, making it more space-efficient than the general binary tree LCA algorithm. Also, clarify that the BST property allows early termination once the split point is found.
State that the LCA is the lowest node in the tree that has both target nodes as descendants (allowing a node to be a descendant of itself).
Explain that for any node, if both target values are less than the node's value, the LCA must be in the left subtree; if both are greater, it must be in the right subtree.
The LCA is the first node encountered where the target values lie on different sides (or one equals the node's value), meaning the node's value is between the two target values.
Start at the root and traverse: if both targets are less, go left; if both are greater, go right; otherwise, return the current node as the LCA.
Conclude that the time complexity is O(h) where h is the tree height (O(log n) for balanced BST, O(n) worst-case), and space is O(1) for iterative traversal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Said you could collect ancestors of p into a set by walking up to root, then walk up from q and return the first node in that set.
Explain that with parent pointers, you can compute the LCA by finding the intersection of the paths from each node to the root. Use a hash set to store ancestors of one node, then traverse from the other node until a common ancestor is found. This yields O(h) time and O(h) space, where h is the height of the tree.
Pro tip: Mention that you can optimize space to O(1) by first computing the depths of both nodes, aligning them, and then moving both pointers upward in tandem until they meet. This demonstrates awareness of trade-offs and shows you can adapt the solution based on constraints.
Confirm that each node has a parent pointer, the tree is not necessarily binary, and nodes are guaranteed to be in the tree. Ask if the tree can be deep (to discuss recursion vs iteration).
Traverse from node A to the root, storing each visited node in a hash set. Then traverse from node B upward until you find a node already in the set; that node is the LCA.
State that time complexity is O(h) where h is the height of the tree, and space complexity is O(h) for the hash set. Note that this is efficient compared to the O(n) traversal needed without parent pointers.
Explain that you can compute the depth of each node by traversing to the root, then move the deeper node up by the depth difference, and finally move both nodes up simultaneously until they meet. This uses constant extra space.
Mention handling cases where one node is an ancestor of the other, and compare the two approaches: hash set is simpler but uses O(h) space; depth alignment is more space-efficient but requires two passes to compute depths.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem constraints: tree size, memory limits, I/O characteristics, and query patterns (preprocessing vs. online). Then propose a disk-based solution using an external memory data structure like a B-tree or LSM-tree to store node metadata (parent pointers, depths, Euler tour), and design an algorithm that minimizes random I/O by leveraging blocking and sequential access. Finally, discuss trade-offs between preprocessing time, query latency, and storage overhead.
Pro tip: Emphasize that you would first check if the tree can be compressed or if LCA can be computed with a streaming algorithm to avoid storing the entire tree. Also, mention that you would consider using a distributed system like MapReduce for preprocessing if the tree is extremely large, and use caching for frequently accessed nodes to reduce I/O.
Ask about tree size, available memory, disk I/O speed, query frequency, and whether preprocessing is allowed. Determine if the tree is static or dynamic.
Decide how to store the tree on disk: e.g., parent pointers, depth, Euler tour, or binary lifting table. Use a layout that supports efficient sequential reads and minimizes random seeks.
If preprocessing is allowed, compute and store necessary data (e.g., Euler tour, sparse table, or binary lifting) using external sorting or MapReduce to handle large data. Optimize for block transfers.
For each query, fetch required nodes from disk using index structures (e.g., B-tree) and compute LCA with minimal I/O. Use caching and prefetching to reduce latency.
Compare approaches (e.g., binary lifting vs. Euler tour + RMQ) in terms of preprocessing time, query time, and I/O cost. Discuss blocking, indexing, and potential parallelism.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.