PAT A-1119 Pre- and Post-order Traversals (30 points)

Topic: 1119 Pre- and Post-order Traversals (30 points)

Analysis : Binary tree, whether the pre-order and post-order are unique, and output in the order

#include <iostream>
#include <vector>
#include <string.h>
#include<stdio.h>
using namespace std;
typedef struct node{
    
    
    int left = -1;
    int right = -1;
};
node node[31];
int pre[31];
int post[31];
int is_u = 1;
void build(int prel,int prer,int postl,int postr)
{
    
    
    if(prel>=prer)
        return ;
    int val = pre[prel+1];
    int index;
    for(int i = postl;i<=postr;i++)
    {
    
    
        if(val == post[i])
        {
    
    
            index = i;
            break;
        }
    }
    int num = index - postl;
    if(num == prer-prel-1)
        is_u = 0;
    node[prel].left = prel+1; // 不唯一  当做左孩子处理
    build(prel+1,prel+1+num,postl,index);
    if(prer-prel-1 != num) //唯一  有右孩子
    {
    
    
        node[prel].right = prel+1+num+1;
        build(prel+1+num+1,prer,index+1,postr);
    }
}
int first = 1;
void inorder(int root)
{
    
    
    if(root==-1)return ;
    inorder(node[root].left);
    if(first)
    {
    
    
        first = 0;
        printf("%d",pre[root]);
    }else
    {
    
    
        printf(" %d",pre[root]);
    }
    inorder(node[root].right);
}
int main()
{
    
    
    int n;
    scanf("%d",&n);
    for(int i = 1;i<=n;i++)
        scanf("%d",&pre[i]);
    for(int i = 1;i<=n;i++)
        scanf("%d",&post[i]);
    build(1,n,1,n);
    if(is_u)
        printf("Yes\n");
    else
        printf("No\n");
    inorder(1);
    printf("\n");
    return 0;
}