返回路径相关类型

如何设计返回路径相关类型的方法？在下面的例子中，我故意想要Vertex路径依赖于Tree这样就禁止跨树混合顶点（这只是一个例子）：

trait Tree {
  trait Vertex
  def root: Vertex
  def addChild(parent: Vertex): Vertex
}

trait TreeFactory { def make: Tree }

现在无法构建以下内容：

def test(f: TreeFactory): (Tree, Map[Tree#Vertex, Tree#Vertex]) = {
  val t  = f.make
  var sq = IndexedSeq(t.root)
  var m  = Map.empty[t.Vertex, t.Vertex]
  for( i <- 1 to 100) {
    val p = sq(util.Random.nextInt(sq.size))
    val c = t.addChild(p)
    m    += c -> p
    sq  :+= c
  }
  (t, m)
}

因为显然我返回的地图不应该有类型的键和值Tree#Vertex但是路径相关的顶点......

error: type mismatch;
 found   : scala.collection.immutable.Map[t.Vertex,t.Vertex]
 required: Map[Tree#Vertex,Tree#Vertex]
Note: t.Vertex <: Tree#Vertex, but trait Map is invariant in type A.
You may wish to investigate a wildcard type such as `_ <: Tree#Vertex`. (SLS 3.2.10)
           (t, m)
               ^

如果我尝试将树创建和父子映射构建解耦：

def test(t: Tree): Map[t.Vertex, t.Vertex] = {
  var sq = IndexedSeq(t.root)
  var m  = Map.empty[t.Vertex, t.Vertex]
  for (i <- 1 to 100) {
    val p = sq(util.Random.nextInt(sq.size))
    val c = t.addChild(p)
    m    += c -> p
    sq  :+= c
  }
  m
}

失败还有另一个原因："error: illegal dependent method type"

我扭曲的头脑想出了这个。我希望有一个更优雅的解决方案：

trait Gagaism {
  val tree: Tree
  val map: Map[tree.Vertex, tree.Vertex]
}

def test(f: TreeFactory) = new Gagaism {
  val tree = f.make
  val map = {
    var sq = IndexedSeq(tree.root)
    var m = Map.empty[tree.Vertex, tree.Vertex]
    for (i <- 1 to 100) {
      val p = sq(util.Random.nextInt(sq.size))
      val c = tree.addChild(p)
      m    += c -> p
      sq  :+= c
    }
    m
  }
}