Title Meaning

Give you an integer n, then an integer n, now let you judge whether the given number is the sequence of a binary search tree or its image.

Solving problems

According to the properties of binary search tree, binary tree can be divided into left and right parts by root, otherwise it is not a standard sequence.
In fact, the image binary tree is the left small right big reverse judgment.

Code part

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=1100;
bool vis;
int pre[maxn];
vector <int> be;
void f(int left,int right)
{
    if(left>right)
        return;
    int tl=left+1;
    int tr=right;
    if(!vis)
    {
        while(tl<=right&&pre[tl]<pre[left])
            tl++;
        while(tr>left&&pre[tr]>=pre[left])
            tr--;
    }
    else
    {
       while(tl<=right&&pre[tl]>=pre[left])
            tl++;
        while(tr>left&&pre[tr]<pre[left])
            tr--;
    }
    if(tl-tr!=1)
        return;
    f(left+1,tr);
    f(tl,right);
    be.push_back(pre[left]);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; ++i)
        {
            scanf("%d",&pre[i]);
        }
        f(0,n-1);
        if(be.size()!=n)
        {
            vis=1;
            be.clear();
            f(0,n-1);
        }
        if(be.size()!=n)
        {
            printf("NO\n");

        }
        else
        {
            printf("YES\n%d",be[0]);
            for(int i=1; i<n; ++i)
                printf(" %d",be[i]);
            printf("\n");
        }
    }
    return 0;
}