I knew BFS was the right move but fumbled on how to track column indices cleanly.
Use BFS with a queue storing nodes along with their column indices, and a map from column index to list of node values. Process level by level to ensure top-to-bottom order, and within each level, process nodes left to right to maintain left-to-right order for same row and column.
Pro tip: Mention that you can avoid sorting by using a TreeMap or by tracking min and max column indices, and clarify that for same row and column, left-to-right order is naturally preserved if you process nodes in level order from left to right.
Confirm that vertical order means grouping nodes by their horizontal distance from the root, and that within each column, nodes are ordered by row (top to bottom) and then left to right.
Use a queue for BFS, storing each node with its column index. Use a hash map (or TreeMap) to map column indices to lists of node values.
Perform BFS starting from the root with column index 0. For each node, add its value to the list for its column, then enqueue its left child with column-1 and right child with column+1.
After traversal, extract the lists from the map in sorted order of column indices (either by sorting keys or using a TreeMap).
Consider empty tree, single node, and skewed trees. Ensure the solution works when multiple nodes share the same row and column.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.