The tricky part is handling ties correctly.
Use a BFS traversal while tracking each node's column index, storing nodes in a map from column to list of values. Then output the lists in sorted column order, ensuring that nodes with the same row and column are ordered left to right by processing left child before right child.
Pro tip: Clarify the tie-breaking rule: when multiple nodes share the same row and column, they should be ordered left to right. This can happen if the tree has nodes that are not in a standard binary tree structure, but in a binary tree, nodes at the same row and column are typically unique. However, if duplicates occur, BFS with left-to-right processing naturally handles it.
Confirm the definition of vertical order: columns from leftmost to rightmost, and within each column, nodes sorted by row (top to bottom). Ask about tie-breaking for same row and column, and discuss handling of empty tree.
Use a queue for level-order traversal, storing each node with its column index. Start with root at column 0. For each node, add its value to a map keyed by column, then enqueue left child with column-1 and right child with column+1.
After traversal, extract the column keys, sort them, and for each column, append its list of values to the result. Since BFS processes nodes level by level, within each column the values are already in top-to-bottom order.
Time complexity is O(n log n) due to sorting columns, but can be O(n) if using a TreeMap or if columns are within a known range. Space complexity is O(n) for the map and queue. Discuss trade-offs.
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