The column grouping part I got pretty quickly with a BFS and tracking offsets.
Use a BFS traversal to process nodes level by level, tracking each node's column index. Store nodes in a map keyed by column, and within each column, maintain a list sorted by row and then by value. Finally, extract the columns in sorted order and return the grouped nodes.
Pro tip: Clarify the tie-breaking rule upfront: if two nodes share the same row and column, sort by value. Also, mention that you can use a TreeMap to keep columns sorted, but if columns are sparse, a hash map with sorting at the end may be more efficient.
Restate the problem to ensure clarity: group nodes by column, order top-to-bottom by row, and sort by value for ties. Consider edge cases like empty tree, single node, and nodes with same row/column.
Use BFS (level-order) to naturally process nodes top-to-bottom. Use a hash map (or TreeMap) to map column indices to a list of (row, value) pairs, and a queue for BFS.
During BFS, for each node, record its column and row (level). Add the node's value to the map under its column, storing row and value. For left child, column-1; for right child, column+1.
After traversal, for each column, sort the list of (row, value) pairs by row, then by value. Then collect the values in order.
Extract columns in sorted order (if using hash map, sort keys) and return a list of lists, each containing the node values for that column.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.