Why is the field node.value and the empty tree null?

The harness gives Scala the node class practice sites usually hand Scala solvers: class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) with var value, var left and var right. A missing child is null, not None, so every read is guarded with != null. If you prefer options, Option(node.left) turns a null into None.

Is it safe to change the queue inside Array.fill?

Yes. Array.fill(n)(elem) evaluates n once and then the by-name elem exactly n times, in index order. Enqueueing children during those evaluations does not change n, so every row stops at the level boundary and [6, 2, 9, 1, 4, null, 12] gives [[6], [2, 9], [1, 4, 12]].

Is there a more functional way to write the BFS?

Yes, with no mutable state: Iterator.iterate(List(root).filter(_ != null))(_.flatMap(n => List(n.left, n.right).filter(_ != null))).takeWhile(_.nonEmpty).map(_.map(_.value).toArray).toArray. Each step maps a level to the next one, and takeWhile stops at the first empty level. It allocates a new List per level, which is fine at this size.

Why does walk return Unit and not the levels?

It only appends to the ArrayBuffer it closes over, so there is nothing to return. An if without an else already has type Unit, and the explicit : Unit makes that intent visible and lets the recursive calls type-check without inference. A pure version would thread the levels through and return them, for example def walk(node: TreeNode, depth: Int, acc: Vector[Vector[Int]]): Vector[Vector[Int]], at the cost of rebuilding a row on every append.

Can the recursive walk overflow the stack?

On a very deep tree, yes. The two recursive calls are not in tail position, so @tailrec cannot apply, and each depth adds a JVM frame; a chain of many thousands of nodes leaning one way ends in StackOverflowError. The mutable.Queue version keeps its nodes on the heap. The cases here stay within 200 levels, so both pass.

Binary Tree Level Order Traversal in Scala: DFS and BFS Solutions

level_order.scala

Scala 3.6

Not run yet

+ 7 hidden

Input

root = [6, 2, 9, 1, 4, null, 12]

Expected

[[6], [2, 9], [1, 4, 12]]

The root 6 is a level of its own, its children 2 and 9 form the next one, and the grandchildren 1, 4 and 12 the last. 9 has no left child, so the third level simply has three values.

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level

1 · Depth-first: file each value under its depth

2 · Breadth-first: one queue round per level