The core idea clicked fast for me, BFS with coordinate tracking.
Use a BFS traversal while tracking each node's column index (root at 0, left child -1, right child +1). Store nodes in a hash map keyed by column, where each value is a list of (row, value) pairs, then sort each column by row and value to produce the final output.
Pro tip: Mention that BFS naturally processes nodes top-to-bottom and left-to-right, so if you append to each column's list in BFS order, you only need to sort by row for nodes at the same position—this avoids a full sort and shows you understand the traversal's ordering guarantees.
Confirm that columns are indexed relative to the root (root at 0, left negative, right positive) and that within a column, nodes are ordered by row first, then by value if they share the same row and column.
Use BFS with a queue storing (node, row, col). Use a hash map mapping col -> list of (row, value) pairs. Track min and max column to avoid sorting keys later.
Process the queue: for each node, append (row, value) to the list for its column. Enqueue left child with (row+1, col-1) and right child with (row+1, col+1).
For each column from min to max, sort its list by row (and value if needed), then extract the values. Return the list of columns.
State time complexity: O(N log N) worst-case due to sorting within columns, but often closer to O(N) if columns are small. Space complexity: O(N) for the map and queue.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.