Use a BFS traversal while tracking the horizontal distance (column) of each node, storing nodes in a map from column to list of node values. Then sort the columns and within each column, ensure nodes are ordered top-to-bottom (BFS naturally gives this if we process level by level).
Pro tip: Clarify the ordering of nodes within the same column and row: if two nodes share the same column and row, they should be ordered left-to-right. Mention that BFS with level-order processing handles this correctly.
Ask about ordering within columns (top-to-bottom, and left-to-right for same row/column) and confirm output format. Discuss edge cases like empty tree, single node, skewed tree.
Decide on BFS (level-order) to ensure top-to-bottom ordering. Use a hash map (or TreeMap) to group nodes by column index, and a queue for BFS that stores nodes along with their column index.
Initialize queue with root at column 0. While queue not empty, process nodes level by level: for each node, add its value to the map entry for its column, then enqueue left child with column-1 and right child with column+1.
After BFS, extract keys (columns) from the map, sort them in ascending order, and for each column, append its list of values to the result. If using TreeMap, iteration is already sorted.
State time complexity O(N log N) due to sorting columns (or O(N) with TreeMap if columns are bounded), space O(N). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The off-by-one adjustment from edges to nodes sounds trivial and it kind of is, but I fumbled the wording when explaining it.
Use a post-order DFS that returns the height (in nodes) of each subtree while updating a global maximum with the sum of left and right heights plus one. Clarify that the diameter is measured in nodes, so the base case for a null node returns 0 and the path length through a node is leftHeight + rightHeight + 1.
Pro tip: Explicitly state that the diameter may not pass through the root, and that the global maximum must be updated at every node. Also mention that the algorithm runs in O(n) time and O(h) space, which is optimal for this problem.
Confirm that the diameter is the number of nodes on the longest path between any two nodes, and that the path may or may not pass through the root.
Design a helper function that returns the height of a subtree in terms of nodes (null returns 0). At each node, compute the left and right heights.
At each node, the longest path through that node has length leftHeight + rightHeight + 1. Update a global variable with the maximum of this value.
Return 1 + max(leftHeight, rightHeight) to the parent, representing the height of the current subtree in nodes.
State that the algorithm visits each node once, giving O(n) time, and uses O(h) space for the recursion stack, where h is the tree height.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.