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Pre- and Post-order Traversals

PAT-1119

• 这题难度较大，主要需要考虑如何实现根据前序遍历和后序遍历来确定一颗二叉树
• 一篇好的文章: 题解

import java.util.Scanner;

/**
 * @Author WaleGarrett
 * @Date 2020/9/5 18:04
 */
public class PAT_1119 {
    static int[] preorder;
    static int[] postorder;
    static boolean flag=true;//判断二叉树是否唯一
    public static void main(String[] args) {
        Scanner scanner=new Scanner(System.in);
        int n=scanner.nextInt();
        preorder=new int[n];
        postorder=new int[n];
        for(int i=0;i<n;i++){
            preorder[i]=scanner.nextInt();
        }
        for(int i=0;i<n;i++){
            postorder[i]=scanner.nextInt();
        }
        PPNode root=buildTree(0,n-1,0,n-1);
        if(flag){
            System.out.println("Yes");
        }else System.out.println("No");
        System.out.println(inOrder(root,"").trim());
    }
    public static PPNode buildTree(int prel,int prer,int postl,int postr){
        PPNode root=new PPNode();
        root.value=preorder[prel];
        if(prel>=prer)
            return root;
//        System.out.println(prel+" "+prer);
        int position=0;
        if(prel<prer&&preorder[prel+1]==postorder[postr-1]){
            flag=false;//不唯一
            //默认为左子树
            root.left=buildTree(prel+1,prer,postl,postr-1);
        }else{
            for(int i=postl;i<postr;i++){
//                System.out.println(i+" "+(prel+1));
                if(postorder[i]==preorder[prel+1]){
                    position=i;
                    break;
                }
            }
            int numl=position-postl+1;
            root.left=buildTree(prel+1,prel+numl,postl,position);
            root.right=buildTree(prel+numl+1,prer,position+1,postr-1);
        }
        return root;
    }
    public static String inOrder(PPNode root,String result){
        if(root.left!=null)
            result=inOrder(root.left,result);
        result=result+root.value+" ";
        if(root.right!=null){
            result=inOrder(root.right,result);
        }
        return result;
    }
}
class PPNode{
    PPNode left;
    PPNode right;
    int value;
    public PPNode(){
        left=right=null;
        value=-1;
    }
}

PAT-1119（Pre- and Post-order Traversals）+前序和后序遍历确定二叉树+判断二叉树是否唯一

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原文地址：https://www.cnblogs.com/GarrettWale/p/13619434.html