What you see from the right of a tree

Walk the tree row by row with a queue and record the last value in each row. Or run a DFS that visits right children first and records the first node seen at each new depth. Both are O(n) time; the DFS uses O(h) stack.

The DFS version uses O(h) space; the BFS version O(w). Mention the trade-off.

A node on the left can be visible if the right side is shorter.

Left view: take the first value of each row instead.

def right_view(root: Node | None) -> list[int]:
    seen: list[int] = []

    def walk(node: Node | None, depth: int) -> None:
        if node is None:
            return
        if depth == len(seen):
            seen.append(node.val)
        walk(node.right, depth + 1)
        walk(node.left, depth + 1)
    walk(root, 0)
    return seen

Common mistakes

• Only following right children from the root, which misses left nodes in deeper rows.
• Recording a value per node instead of per row.

Draw a tree whose right side is short to show why 'follow right children' fails.

Follow-up questions

Think of your answer first, then open the hint.

1. What about the view from the bottom?

   Show hint

   Track horizontal positions (column) with BFS and keep the last value per column.

2. Can you do it without a queue?

   Show hint

   Yes: DFS right-first, record when depth == len(result).