BFS without queue

bfs.fsx

type 'a Tree =
    | Leaf
    | Node of 'a * 'a Tree * 'a Tree

let bfs (tree: 'a Tree) =
    [ tree ]
    |> Seq.unfold
        (function
        | [] -> None
        | h :: t ->
            match h with
            | Leaf -> Some(None, t)
            | Node (x, l, r) -> Some(Some(x), t @ [ l; r ]))
    |> Seq.filter Option.isSome
    |> Seq.map Option.get


let t =
    Node(0, Node(1, Node(3, Leaf, Leaf), Leaf), Node(2, Leaf, Node(4, Leaf, Leaf)))

bfs t |> printfn "%A"
//seq [0; 1; 2; 3; 4]