Bytedance·Backend Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Bytedance backend round, one algorithmic problem the whole time. Pretty standard tree recursion territory but the edge cases tripped me up more than I expected.

Questions Asked (1)

Given two binary trees A and B, determine whether B is a substructure of A. B is a substructure if there exists a node in A where, starting from that node, every node in B matches the corresponding node in A by value and position. B does not need to reach the leaves of A, and a null B is never considered a substructure.

Algorithms & Data Structures

Author's notes

My first instinct was to just check if B is a subtree, which is wrong.

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AI HintsAI Generated

Suggested Approach

Use a recursive approach: traverse tree A, and for each node, check if the subtree rooted at that node is identical to B using a helper function. The helper compares nodes recursively, ensuring values match and structure aligns, returning true if B is fully matched. If any node in A yields a match, B is a substructure.

Pro tip: Clarify edge cases upfront: B null is never a substructure, but A null with non-null B returns false. Also, discuss time complexity: O(m*n) worst-case, but can be optimized with tree hashing or serialization if needed.

1. Clarify problem and edge cases

Confirm definitions: B is a substructure if there's a node in A where B matches exactly from that node downward. Discuss edge cases: B null -> false; A null -> false if B non-null; single-node trees.

2. Design recursive solution

Define a helper function isSameTree(nodeA, nodeB) that returns true if the subtree rooted at nodeA is identical to nodeB. Then, traverse A: for each node, if isSameTree(node, B) is true, return true; otherwise recurse on left and right children.

3. Analyze complexity and optimizations

Explain worst-case time complexity O(m*n) where m and n are sizes of A and B, due to repeated comparisons. Mention potential optimizations like tree hashing or serialization to reduce to O(m+n) but note trade-offs.

4. Implement and test

Write clean code with base cases: if B is null, return false; if A is null, return false. Test with examples: A=[1,2,3], B=[2] -> true; A=[1,2,3], B=[1,2] -> true; A=[1,2,3], B=[1,2,4] -> false.

Key Points to Mention

Recursive traversal of tree A and comparison with B at each node.
Helper function to check if two subtrees are identical (value and structure).
Edge cases: null B, null A, and B larger than A.
Time complexity O(m*n) and space complexity O(h) for recursion stack.
Potential optimizations like tree hashing or serialization for better performance.
Clarify that B does not need to reach leaves of A; it can be a partial subtree.

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