Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Bytedance SWE coding round, one problem, and it did not go great. Needed hints to get through it and left feeling like I'd underprepped the tree stuff.

Questions Asked (1)

Given a binary tree where each node has a pointer to its parent, find the lowest common ancestor of two given nodes.

Algorithms & Data Structures

Author's notes

I knew the general LCA problem but had not worked through the version with parent pointers.

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AI HintsAI Generated

Suggested Approach

Since each node has a parent pointer, you can find the LCA by first computing the depths of both nodes, then moving the deeper node up to the same depth, and finally moving both nodes up in tandem until they meet. This approach runs in O(h) time and O(1) space, where h is the height of the tree.

Pro tip: Mention that this parent-pointer approach is more efficient than the standard recursive method for this specific tree structure, and discuss edge cases like when one node is an ancestor of the other or when nodes are in different trees.

1. Clarify assumptions and edge cases

Confirm that the tree is binary, nodes have parent pointers, and both nodes are in the same tree. Ask about handling cases where one node is the ancestor of the other or if nodes could be null.

2. Compute depths of both nodes

Traverse from each node up to the root using parent pointers to determine the depth (distance from root) of each node. This can be done by counting steps until the root is reached.

3. Align depths

If the depths differ, move the deeper node up by the difference in depths so that both nodes are at the same level.

4. Find the LCA

Move both nodes up simultaneously (using parent pointers) until they point to the same node. That node is the lowest common ancestor.

5. Analyze complexity and discuss alternatives

State that the time complexity is O(h) and space is O(1). Briefly mention alternative approaches like using a hash set to store ancestors of one node, which takes O(h) space.

Key Points to Mention

Time complexity: O(h) where h is the height of the tree, and space complexity: O(1) since no extra data structures are used.
Handling the case where one node is an ancestor of the other: the algorithm naturally returns the ancestor node.
The importance of parent pointers: without them, you'd need to traverse from the root, which could be less efficient.
Edge cases: null nodes, nodes not in the same tree, or the tree being a single node.
Comparison with the recursive approach for trees without parent pointers, which typically uses O(h) space for the call stack.
Potential follow-up: if parent pointers are not available, how would you solve it? (e.g., using a hash set to store ancestors).

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