The BFS part clicked pretty fast for me, the part I had to think through was how to assign column indices cleanly.
Use a BFS level-order traversal while tracking each node's column index, storing nodes in a hash map keyed by column. After traversal, sort the columns and output the values in order, ensuring that within each column nodes are ordered by level and then by left-to-right order.
Pro tip: Clarify the tie-breaking rule for nodes in the same column and level: they should appear in the order they are visited during level-order traversal (i.e., left-to-right). Also, mention that using a TreeMap can avoid a separate sorting step, but be prepared to discuss the trade-offs.
Confirm the definition of vertical order: columns are indexed from leftmost (smallest) to rightmost (largest). Within each column, nodes are ordered top-to-bottom, and for nodes at the same level, left-to-right as encountered in level-order traversal.
Use a queue for BFS, storing each node along with its column index. Use a hash map (or TreeMap) to group node values by column index, preserving insertion order within each column.
Perform a level-order traversal starting with the root at column 0. For each node, append 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 column indices, sort them, and for each column in sorted order, output the list of node values.
State that the time complexity is O(n log n) due to sorting columns (or O(n) if using a TreeMap with O(log n) insertion per node, but overall O(n log n)), and space complexity is 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.