Scala - Check If object exist inside a list -

i'm trying write algorithm check if object exists inside list.

my case class following:

case class person(id:int, name:string, friends:list[person] = nil) 

i've wrote using code:

@tailrec final def find(id: int, tree: list[person]):option[person] = tree match {   case nil => none   case (h :: t) => if(h.id == id) option(h) else find(key, t ::: h.friends) } 

is approach? use tail recursive , append list on tail list? if not, best approach?

i ended following implementation tail recursive bfs:

@tailrec def find(id: int, level: list[person], visited: set[person] = set.empty): option[person] =   if (level.isempty) none   else {     // try find person on level     val found = level.find(_.id == id).filternot(visited.contains)     if (found.isdefined) found     else find(       id,       // next level construction       level.flatmap(_.friends).distinct,       // keep track of visited handle cycles       visited ++ level     )   } 

update: also, suggest add call name able play around more test cases:

class person(val id: int, val name: string, friends: => list[person]) {    def friendlist = friends    override def tostring = name }  object person {    def apply(id: int, name: string, friends: => list[person] = nil) = new person(id, name, friends) } 

in case can compose example comments:

val john = person(1, "john") val russ = person(2, "russ") val bob = person(3, "bob") val lois = person(5, "lois")  lazy val `eve friends` = list(john, anne) lazy val `peter friends` = list(eve, bob, russ) lazy val `anne friends` = list(peter, lois)  val eve = person(7, "eve", `eve friends`) val peter = person(8, "peter", `peter friends`) val anne: person = person(6, "anne", `anne friends`)  println(find(8, list(eve, peter, lois))) // result: some(peter)