Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Meta SWE interview with a tree problem that sounds straightforward until you realize the root isn't given to you. The parent pointer twist changes everything about how you approach it.

Questions Asked (1)

Given two nodes p and q in a binary tree where every node has a parent pointer, find their lowest common ancestor. You are not given the root directly and must navigate using parent and child pointers.

Algorithms & Data Structures

Author's notes

My first instinct was to just walk both nodes up to the root and collect the paths, then compare.

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AI HintsAI Generated

Suggested Approach

Since parent pointers are available, treat the problem as finding the intersection of two linked lists: compute the depth of each node by walking up to the root, align the deeper node, then move both pointers up in tandem until they meet. This yields O(h) time and O(1) space, where h is the tree height.

Pro tip: Mention that you can avoid computing depths by using a two-pointer technique where each pointer traverses up to the root and then switches to the other node's path; they will meet at the LCA after at most two passes. This demonstrates deeper insight and often impresses interviewers.

1. Clarify assumptions and constraints

Confirm that parent pointers are valid, nodes are in the same tree, and p and q are distinct. Ask about edge cases like one node being the ancestor of the other.

2. Compute depths

Write a helper function to find the depth of a node by following parent pointers until null. Compute depths for both p and q.

3. Align depths

Move the deeper node up by the difference in depths so that both pointers are at the same level.

4. Find intersection

Move both pointers up simultaneously until they point to the same node. That node is the lowest common ancestor.

5. Analyze complexity and edge cases

State that time complexity is O(h) and space is O(1). Discuss edge cases: p or q is the LCA, nodes at different depths, and tree with only one node.

Key Points to Mention

Time complexity O(h) and space complexity O(1), where h is the height of the tree.
Handling the case where one node is an ancestor of the other (the LCA is that node).
The two-pointer technique that avoids explicit depth calculation by switching paths.
Edge cases: p and q are the same node, tree with only one node, or nodes not in the same tree (if not guaranteed).
Comparison to the classic LCA problem without parent pointers (which requires recursion or storing paths).
Potential follow-up: what if the tree is very deep and recursion is not allowed? (Iterative solution is already used.)

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.