Is It a Binary Search Tree

PAT-1043

• 主要涉及到根据前序遍历序列片段是否是一颗二叉树，这里有一个小tip就是插入序列就是二叉树的前序遍历序列。
• 第二个是会对排序二叉树进行前序遍历，后序遍历，镜像排序二叉树的前序遍历，后序遍历等操作。
• 题目的整体难度不大，但是对二叉树和二叉排序树的了解需要较深。

/**
 * @Author WaleGarrett
 * @Date 2020/9/5 8:20
 */

import java.util.Scanner;

/**
 * PAT-1043,1064,1066,1086,1089,1099,1098
 * 7
 * 8 6 5 7 10 8 11
 */
public class PAT_1043 {
    static final int maxn=1003;

    public static void main(String[] args) {
        BinaryTree binaryTree=new BinaryTree();
        Scanner scanner=new Scanner(System.in);
        String result="";
        int n=scanner.nextInt();
        while(n!=0){
            int value=scanner.nextInt();
            binaryTree.insert(value);//插入输入串
            result=result+value+" ";
            n--;
        }
        //使用前序遍历该构建完成的排序二叉树，并且对该树的镜像二叉树进行前序遍历，任何一个序列和题目的序列匹配则说明符合要求，再输出该排序二叉树的后序遍历
        String preorder=binaryTree.preOrder(binaryTree.root,"");//前序遍历
        String mpreorder=binaryTree.mPreOrder(binaryTree.root,"");//镜像前序遍历
        if(preorder.equals(result)){
            System.out.println("YES");
            System.out.println(binaryTree.postOrder(binaryTree.root,"").trim());
        }else if(mpreorder.equals(result)){
            System.out.println("YES");
            System.out.println(binaryTree.mPostOrder(binaryTree.root,"").trim());
        }else
            System.out.println("NO");
    }
}
class Node{
    Node left;
    Node right;
    int value;
    public Node(){
        left=right=null;
        value=-1;
    }
    public Node(Node left,Node right,int value){
        this.value=value;
        this.left=left;
        this.right=right;
    }
}
class BinaryTree{
    Node root;
    public BinaryTree(){
        root=null;
    }
    public void insert(int value){
        if(root==null){
            root=new Node();
            root.value=value;
        }else{
            Node now=root;
            while(true){
                if(value<now.value){
                    if(now.left==null){
                        now.left=new Node(null,null,value);
                        break;
                    }else now=now.left;
                }else{
                    if(now.right==null){
                        now.right=new Node(null,null,value);
                        break;
                    }else{
                        now=now.right;
                    }
                }
            }
        }
    }
    public String preOrder(Node now,String result){
        result=result+now.value+" ";
        Node left=now.left;
        if(left!=null){
            result=preOrder(left,result);
        }
        Node right=now.right;
        if(right!=null){
            result=preOrder(right,result);
        }
        return result;
    }
    public String mPreOrder(Node now,String result){
        result=result+now.value+" ";
        Node right=now.right;
        if(right!=null){
            result=mPreOrder(right,result);
        }
        Node left=now.left;
        if(left!=null){
            result=mPreOrder(left,result);
        }
        return result;
    }
    public String postOrder(Node now,String result){
        Node left=now.left;
        if(left!=null){
            result=postOrder(left,result);
        }
        Node right=now.right;
        if(right!=null){
            result=postOrder(right,result);
        }
        result=result+now.value+" ";
        return result;
    }
    public String mPostOrder(Node now,String result){

        Node right=now.right;
        if(right!=null){
            result=mPostOrder(right,result);
        }
        Node left=now.left;
        if(left!=null){
            result=mPostOrder(left,result);
        }
        result=result+now.value+" ";
        return result;
    }
}