I went with a BFS approach, tracking (node, col, depth) tuples and dumping everything into a map from column to list of (depth, value) pairs.
Start by clarifying the tie-breaking rule and confirming the definition of vertical order. Then, describe a BFS approach using a queue and a map from column to list of values, processing nodes level by level to ensure top-to-bottom and left-to-right ordering. Finally, analyze time and space complexity and discuss potential optimizations.
Pro tip: Mention that using BFS naturally handles the tie-breaking rule because nodes at the same depth are processed left-to-right, and within the same column, their order is preserved. Also, note that a DFS approach would require sorting by depth and horizontal position, which adds complexity.
Restate the problem in your own words, confirm the tie-breaking rule (top-to-bottom, left-to-right within same depth and column), and ask about edge cases (e.g., empty tree, single node).
Decide between BFS and DFS. BFS is preferred because it processes nodes level by level, naturally satisfying the top-to-bottom and left-to-right ordering within each column.
Use a queue for BFS, storing nodes along with their column indices. Use a hash map (or dictionary) to group node values by column, and keep track of the minimum and maximum column indices to output columns in order.
Initialize the queue with the root at column 0. While the queue is not empty, dequeue a node, add its value to the map under its column, and enqueue its left child with column-1 and right child with column+1. After BFS, iterate columns from min to max and output the lists.
Time complexity is O(n) because each node is visited once. Space complexity is O(n) for the queue and the map. Mention that the map's lists preserve insertion order, which matches the required ordering.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is basically an in-order traversal where you're stitching nodes together as you go.
Start by clarifying the problem and edge cases, then present an in-order traversal approach that rewires pointers in-place. Compare recursive and iterative implementations, discussing trade-offs in time/space and handling of skewed trees.
Pro tip: Emphasize that the transformation must be in-place and that the resulting list should be non-circular; explicitly set the head's prev and tail's next to null to avoid cycles.
Confirm that the list should be sorted (in-order), non-circular, and in-place. Discuss edge cases: empty tree, single node, skewed tree (all left or all right).
Decide between recursive and iterative in-order traversal. Recursive is simpler but uses O(h) stack space; iterative uses explicit stack, also O(h) space, but can be more robust for deep trees.
During traversal, maintain a prev pointer. For each node, set node.left = prev, and if prev exists, prev.right = node. Update prev to current node.
The first node visited becomes the head; set its left to null. After traversal, set the last node's right to null to ensure non-circularity.
Time: O(n) for both approaches. Space: O(h) for recursion/stack, where h is tree height; worst-case O(n) for skewed trees. Discuss iterative vs recursive readability and stack overflow risk.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.