Scala需要这个演员吗？

Question

假设我有一个简单abstract BinaryTree的子类Node,Leaf我想写一个产生一个的函数List[Leaf].

def getLeaves(tree: BinaryTree): List[Leaf] =
    tree match {
      case Leaf(v) => List(tree.asInstanceOf[Leaf])
      case Node(left, right) => getLeaves(left) ++ getLeaves(right)
    }

有没有办法避免案件中的明确asInstanceOf[Leaf]演员？如果我把它遗漏,我得到一个诊断说法:发现:BinaryTree; 需要叶.Leaf

Answer 1

我在其他地方看过这个结构.它似乎做了这个工作.

def getLeaves(tree: BinaryTree): List[Leaf] =
    tree match {
      case leaf: Leaf => List(leaf)
      case Node(left, right) => getLeaves(left) ++ getLeaves(right)
    }