#199 - Binary Tree Right Side View

MediumBinary Tree - BFS

Given the root of a binary tree, return the values of the nodes you can see when looking at the tree from the right side, ordered from top to bottom.

Input	Output
root = [1,2,3,null,5,null,4]	[1,3,4]
root = [1,null,3]	[1,3]
root = []	[]

Solution Space

Try it

/**
 * Returns the rightmost node value at each level (right side view).
 *
 * @param root Root of the binary tree
 * @returns Array of rightmost values from top to bottom
 */
export function rightSideView(root: TreeNode | null): number[] {
  if (!root) return [];
  const result: number[] = [];
  const queue: TreeNode[] = [root];

  while (queue.length > 0) {
    const levelSize = queue.length;

    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift()!;

      // Last node in this level is visible from the right
      if (i === levelSize - 1) result.push(node.val);

      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }

  return result;
}

Approach: BFS level-order traversal – process each level; the last node in each level is the one visible from the right.

Time: O(n) – every node is enqueued and dequeued once.

Space: O(n) – the queue can hold up to n/2 nodes (last level of a complete tree).

Alternative approaches:

Approach	Idea	Time	Space
DFS (right-first)	Traverse right child before left, recording the first node encountered at each depth	O(n)	O(h)

Why BFS is preferred: the level-by-level structure makes the "last in level" logic intuitive; DFS requires tracking depth explicitly.

#1448 Count Good Nodes in Binary Tree

#1161 Maximum Level Sum of a Binary Tree