9-二叉树-Scala实现

Posted 2020-12-14 help-silence

构建二叉树；实现前序、中序、后序遍历；两种删除节点的原则

package com.atguigu.datastructures.binarytree

object BinaryTreeDemo {
  def main(args: Array[String]): Unit = {
    //先使用比较简单的方法，直接关联
    val root = new HeroNode(1,"宋江")
    val hero2 = new HeroNode(2,"吴用")
    val hero3 = new HeroNode(3,"卢俊义")
    val hero4 = new HeroNode(4,"林冲")
    val hero5 = new HeroNode(5,"关胜")
    root.left = hero2
    root.right = hero3
    hero3.left = hero5
    hero3.right = hero4
    val binaryTree = new BinaryTree
    binaryTree.root = root
  //  binaryTree.preOrder()
  //  binaryTree.infixOrder()
  //  binaryTree.postOrder()
//    val resNode: HeroNode = binaryTree.preOrderSearch(5)
//    if (resNode!=null){
//      printf("找到，编号=%d name=%s",resNode.no,resNode.name)
//    }else{
//      printf("没有找到")
//    }
//  val resNode: HeroNode = binaryTree.postOrderSearch(5)
//    if (resNode!=null){
//      printf("找到，编号=%d name=%s",resNode.no,resNode.name)
//    }else{
//      printf("没有找到")
//    }
    binaryTree.delNode2(3)
    binaryTree.preOrder()
  }
}
//定义节点
class HeroNode(hNo:Int,hName:String){
  val no= hNo
  var name = hName
  var left:HeroNode = null
  var right:HeroNode = null

  //前序查找
  def preOrderSearch(no:Int):HeroNode={
    if (no == this.no){
      return this
    }
    //向左递归查找
    var resNode:HeroNode = null
    if (this.left != null){
      resNode=this.left.preOrderSearch(no)
    }
    if (resNode != null){
      return resNode
    }
    //向右边递归查找
    if (this.right != null){
      resNode = this.right.preOrderSearch(no)
    }
    return resNode
  }
  //前序遍历
  def preOrder(): Unit ={
    //先输出当前节点值
    printf("节点信息 no=%d name=%s
",no,name)
    //向左递归输出左子树
    if (this.left != null){
      this.left.preOrder()
    }
    if (this.right != null){
      this.right.preOrder()
    }
  }
  //中序查找
  def infixOrderSearch(no:Int):HeroNode={
    var resNode:HeroNode = null
    //先向左递归查找
    if (this.left !=null){
      resNode=this.left.infixOrderSearch(no)
    }
    if (resNode != null){
      return resNode
    }
    if (no == this.no){
      return this
    }
    //向右递归查找
    if(this.right != null){
      resNode = this.right.infixOrderSearch(no)
    }
    return resNode
  }
  def infixOrder(): Unit ={
    if (this.left != null){
      this.left.infixOrder()
    }
    printf("节点信息 no=%d name=%s
",no,name)
    if (this.right != null){
      this.right.infixOrder()
    }
  }
  //后序查找
  def postOrderSearch(no:Int):HeroNode={
    var resNode:HeroNode = null
    if (this.left !=null){
      resNode=this.left.postOrderSearch(no)
    }
    if (resNode != null){
      return resNode
    }
    if (this.right != null){
      resNode = this.right.postOrderSearch(no)
    }
    if (resNode != null){
      return resNode
    }
    if (no == this.no){
      return this
    }
     resNode
  }
  def postOrder(): Unit ={
    if (this.left != null){
      this.left.postOrder()
    }
    if (this.right != null){
      this.right.postOrder()
    }
    printf("节点信息 no=%d name=%s
",no,name)
  }

  //删除节点
  //规则：如果是叶子节点，直接删除这个节点；非叶子节点，直接删除该子树
  def delNode(no:Int): Unit ={
    //首先比较当前节点的左子节点是否为要删除的节点
    if (this.left != null && this.left.no == no)
      {

        this.left = null
        return
      }
    //比较当前节点的右子节点是否为要删除的节点
    if (this.right != null && this.right.no == no){

        this.right = null
        return
      }
     //向左递归
    if (this.left != null){
      this.left.delNode(no)
    }
    if (this.right != null){
      this.right.delNode(no)
    }

  }

  /**
    * 另一种删除规则
    * @param no
    */
  def delNode2(no:Int): Unit ={
    if (this.left != null && this.left.no == no)
    {
      if (this.left.left != null && this.left.right != null){
        val temp=this.left.right
        this.left = this.left.left
        this.left.left = temp
        return
      }
      if (this.left.left != null && this.left.right == null){
        val temp= this.left.left
        this.left = temp
        return
      }
      this.left = null
      return
    }
    //比较当前节点的右子节点是否为要删除的节点
    if (this.right != null && this.right.no == no){
      if (this.right.left != null && this.right.right != null){
        val temp=this.right.right
        this.right = this.right.left
        this.right.left = temp
        return
      }
      if (this.right.left != null && this.right.right == null){
        val temp= this.right.left
        this.right = temp
        return
      }
      this.right = null
      return
    }
    //向左递归
    if (this.left != null){
      this.left.delNode(no)
    }
    if (this.right != null){
      this.right.delNode(no)
    }

  }
}

class BinaryTree{
  var root:HeroNode = null

  //前序查找
  def preOrderSearch(no:Int):HeroNode={
    if (root !=null){
      return root.preOrderSearch(no)
    }else{
      return null
    }
  }
  //前序遍历
  def preOrder(): Unit ={
    if (root != null){
      root.preOrder()
    }else{
      println("当前树为空")
    }
  }

  //中序遍历查找
  def infixOrderSearch(no:Int):HeroNode={
    if (root != null){
        root.infixOrderSearch(no)
    }else{
      null
    }
  }
  def infixOrder(): Unit ={
    if (root != null){
      root.infixOrder()
    }else{
      println("当前树为空")
    }
  }
  //后序遍历查找
  def postOrderSearch(no:Int):HeroNode={
    if (root != null){
      root.postOrderSearch(no)
    }else{
      null
    }
  }
  def postOrder(): Unit ={
    if (root != null){
      root.postOrder()
    }else{
      println("当前树为空")
    }
  }


  //删除指定节点
  def delNode(no:Int): Unit ={
    if (root != null){
      if (root.no == no){
        root=null
      }else {
        root.delNode(no)
      }
    }
  }
  def delNode2(no:Int): Unit ={
    if (root != null){
      if (root.no == no){
        root=null
      }else {
        root.delNode2(no)
      }
    }
  }

}