Scala binary tree

Tree.scala

abstract class Tree[+A] {
  def value: Option[A] = this match {
    case n: Node[A] => Some(n.v)
    case l: Leaf[A] => Some(l.v)
    case Empty      => None
  }
  def left: Option[Tree[A]] = this match {
    case n: Node[A] => Some(n.l)
    case l: Leaf[A] => None
    case Empty      => None
  }
  def right: Option[Tree[A]] = this match {
    case n: Node[A] => Some(n.r)
    case l: Leaf[A] => None
    case Empty      => None
  }
  def count: Int = {
    def loop(t: Tree[_]): Int = t match {
      case n: Node[_] => Seq(loop(n.left.get), loop(n.right.get)).sum + 1
      case l: Leaf[_] => 1
      case Empty      => 0
    }
    loop(this)
  }
}
case class Node[A](v: A, l: Tree[A], r: Tree[A]) extends Tree
case class Leaf[A](v: A) extends Tree
case object Empty extends Tree[Nothing]

object Run extends App {
  val t: Tree[Symbol] = Node('A, Node('B, Leaf('D), Leaf('E)), Node('C, Empty, Node('F, Leaf('G), Empty)))
  
  println("tree: " + t)

  //print the value of b node navigating from root
  for {
    b <- t.left
    value <- b.value
  } println("B node: " + value)

  //print the value of e node navigating from root
  for {
    b <- t.left
    e <- b.right
    value <- e.value
  } println("E node: " + value)

  //no println() is executed for empty node chain
  for {
    b <- t.left
    e <- b.right
    x <- e.right  //no node here
    value <- x.value
  } println("X node SHOUL NOT PRINT!: " + value)
  
  
  println("count "  + t.count)
/****** output *********
tree: Node('A,Node('B,Leaf('D),Leaf('E)),Node('C,Empty,Node('F,Leaf('G),Empty)))
B node: 'B
E node: 'E
count 7
*************************/
}

Beautiful.

🎉

Would some one please show my how to use foldLoop or fold to return a modified tree?

Good job!

Exquisite.

very cool, nice work :)