再帰を使って木構造をたどる

traversetree.scala

// 頭が慣れていないので、再帰むずい
case class Tree(id: Int, children: List[Tree])
val n8 = Tree(8, Nil)
val n4 = Tree(4, Nil)
val n5 = Tree(5, Nil)
val n6 = Tree(6, List(n8))
val n7 = Tree(7, Nil)
val n2 = Tree(2, List(n4, n5))
val n3 = Tree(3, List(n6, n7))
val n1 = Tree(1, List(n2, n3))

import scala.annotation.tailrec
def traverseTree(t: Tree): List[Int] = {
  @tailrec
  def _traverseTree(rest: List[Tree], result: List[Int]): List[Int] = {
    rest match {
      case Nil =>
        result.reverse
      case x :: xs =>
        _traverseTree(x.children ::: xs, x.id :: result)
    }
  }
  _traverseTree(List(t), Nil)
}
traverseTree(n1)