scala写算法-从后缀表达式构造

一个例子,比如ab+cde+**,这是一个后缀表达式,那么如何转换为一棵表达式树呢?

先上代码,再解释:

object Main extends App{
  import Tree.node
  def isOperator(char: Char):Boolean="+-*/".contains(char)
  val stack=new Stack[Tree]
  val str="ab+cde+**"
  str.foreach(ch=>{
    isOperator(ch) match {
      case true =>
        val left=stack.pop()
        val right=stack.pop()
        stack.push(node(ch,left,right)) //操作符,弹出两个节点,作为新树的左右节点,把当前的操作符作为root,压栈
      case false=>stack.push(node(ch))//非操作符,则新建节点,压栈
    }
  })
  stack.pop().print
}

sealed trait Tree{self=>
  def print={
    def loop(tree:Tree,depth:Int):Unit=tree match {
      case EmptyNode => EmptyNode
      case Node(value,left,right) =>
        loop(right,depth+1)
        println("  "*depth+value)
        loop(left,depth+1)
    }
    loop(self,1)
  }
}
case class Node(value:Char,left:Tree,right:Tree)extends Tree
case object EmptyNode extends Tree{
  override def toString: String = "N"
}
object Tree{
  def node(value:Char,left:Tree=EmptyNode,right:Tree=EmptyNode):Tree=Node(value,left,right)
  def empty:Tree=EmptyNode
}

class Stack[T](implicit classTag: ClassTag[T]){
  val elements=new Array[T](100)
  private var  size=0
  def isEmpty:Boolean=size==0
  def push(x:T)={
    elements(size)=x
    size+=1
  }
  def pop():T={
    if(isEmpty)
      throw new Exception("empty stack")
    size-=1
    elements(size)
  }
}

依次扫描字段串中的每个字符,如果是字母(非操作符),那么把这个字母作为树的节点(只是单节点)放入stack中;如果是操作符(isOperator),那么从stack中弹出两个节点作为新树的左右节点,依次进行到底.最后stack中只有一棵树,这棵树就是表达式树.