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Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026Remote

Summary

Meta phone screen for a SWE role that leaned heavily on tree traversal problems. The vertical order and right side view questions came up as a combo, which I half-expected but still fumbled parts of.

Questions Asked (3)

Q1

Given a binary tree, return the nodes visible from the right side (right side view).

Algorithms & Data Structures
Author's notes

I went with BFS and grabbed the last node of each level, which worked fine.

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AI HintsAI Generated

Suggested Approach

Use BFS level-order traversal, recording the last node at each level. Alternatively, use DFS prioritizing the right child, tracking depth to capture the first node seen at each new depth. Both approaches yield O(n) time and O(h) space.

Pro tip: Clarify that the right side view is the set of rightmost nodes at each depth, not just nodes on the rightmost path. Mention that BFS is more intuitive but DFS uses less memory for skewed trees.

1. Clarify the problem

Confirm that the right side view includes the rightmost node at each depth, even if it's not on the rightmost path. Discuss edge cases like empty tree or single node.

2. Choose an approach

Decide between BFS (level-order) and DFS (right-first). Explain the trade-offs: BFS is straightforward, DFS can be more space-efficient for deep trees.

3. Implement BFS

Use a queue to process nodes level by level. For each level, record the last node's value. Enqueue left and right children for the next level.

4. Implement DFS (alternative)

Traverse right child first, then left. Keep track of the current depth and add the node's value to the result if it's the first time visiting that depth.

5. Analyze complexity

State that both approaches run in O(n) time and O(h) space, where h is the tree height. For BFS, space is O(w) where w is the maximum width.

Key Points to Mention

  • Level-order traversal (BFS) using a queue
  • DFS with right-first traversal and depth tracking
  • Time complexity O(n) and space complexity O(h) or O(w)
  • Handling edge cases: empty tree, skewed tree
  • Difference between right side view and rightmost path
  • Potential follow-up: return nodes from left side view or both sides

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

Q2

Given a binary tree, return its vertical order traversal grouped by column.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

BFS with a (node, col) queue and a defaultdict felt natural once I started writing it.

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AI HintsAI Generated

Suggested Approach

Use a BFS or DFS traversal while tracking each node's horizontal distance (column) from the root. Store nodes in a hash map keyed by column, then output columns in sorted order, sorting nodes within each column by depth and value as needed.

Pro tip: Clarify upfront whether nodes in the same column should be ordered by depth (top-to-bottom) and then by value, as Meta often expects this tie-breaking. Mention that you can avoid sorting columns by tracking min/max column indices during traversal.

1. Clarify requirements and edge cases

Ask about tie-breaking rules (depth vs. value), whether to include empty columns, and how to handle a null root. Confirm the expected output format.

2. Choose traversal and data structures

Select BFS (level-order) or DFS (preorder) to traverse the tree. Use a hash map mapping column index to a list of nodes, and track min/max column indices.

3. Traverse and record nodes

During traversal, compute each node's column (root = 0, left = parent-1, right = parent+1). Append the node's value to the map entry for that column, along with its depth if needed for sorting.

4. Sort and assemble result

Sort columns from min to max. For each column, sort nodes by depth (and value if required). Collect the values into the final list of lists.

5. Analyze complexity and test

State time and space complexity (O(n log n) if sorting, O(n) with BFS and ordered map). Walk through a small example and edge cases like skewed trees.

Key Points to Mention

  • Horizontal distance concept: root at column 0, left child -1, right child +1.
  • Choice of BFS vs. DFS and how it affects ordering within columns.
  • Use of hash map to group nodes by column, and tracking min/max columns to avoid sorting keys.
  • Tie-breaking rules: nodes in same column ordered by depth (top-to-bottom) and then by value.
  • Time and space complexity: O(n log n) with sorting, O(n) with BFS and ordered map; space O(n).
  • Edge cases: empty tree, single node, skewed tree, and columns with multiple nodes at same depth.

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

Q3

Can you return both the left view and the right view of a binary tree in a single BFS pass?

Algorithms & Data Structures
Author's notes

This was the follow-up that got me.

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AI HintsAI Generated

Suggested Approach

Explain that a single BFS pass can capture both views by tracking the first and last node at each level. Use a queue for level-order traversal and record the first node as the left view and the last node as the right view.

Pro tip: Mention that this approach is optimal in time and space, and clarify that the left view is not simply the left child of each node but the leftmost node at each depth.

1. Clarify the problem

Confirm that the left view consists of the first node at each level and the right view consists of the last node at each level. Discuss edge cases like empty tree or single node.

2. Outline BFS approach

Describe using a queue to perform level-order traversal. For each level, process all nodes, and while processing, note the first and last node encountered.

3. Detail the algorithm

Initialize an empty queue and enqueue the root. While the queue is not empty, determine the level size, then iterate through the level: for the first node, add to left view; for the last node, add to right view. Enqueue children of each node.

4. Analyze complexity

State that time complexity is O(n) since each node is visited once, and space complexity is O(w) where w is the maximum width of the tree, due to the queue.

5. Discuss alternatives and trade-offs

Mention that separate DFS or BFS passes could work but would be less efficient. Emphasize that the single BFS pass is optimal and elegant.

Key Points to Mention

  • Level-order traversal using a queue
  • Tracking first and last node per level
  • Time complexity O(n) and space complexity O(w)
  • Handling edge cases: empty tree, single node, skewed tree
  • Comparison with separate passes for left and right views
  • Definition of left/right view: leftmost/rightmost node at each depth

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