传送门
已知前序，并且因为二叉搜索树的特性，中序也确定，只需判断能不能建树，能就输出后序
思路类似于1027，多了一个镜像树的过程

版本一，照着y总写的，非常的简短风骚

#include <iostream>
#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define scl(x) scanf("%lld",&x)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define prl(x) printf("%lld\n",x)
#define ll long long
#define PII pair<int,int>
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1e5+3;
int in[maxn],pre[maxn],post[maxn],cnt;
bool build(int inL,int inR,int preL,int preR,int type){
    if(inL>inR) return true;
    int root=pre[preL];
    int k;
    if(type==0){
        k=inL;
        while(in[k]!=root&&k<=inR) k++;
        if(k>inR) return false;
    }
    else{
        k=inR;
        while(in[k]!=root&&k>=inL) k--;
        if(k<inL) return false;
    }
    bool res=true;
    if(!build(inL,k-1,preL+1,preL+1+(k-1-inL),type)) res=false;
    if(!build(k+1,inR,preL+1+(k-1-inL)+1,preR,type)) res=false;
    post[cnt++]=root;
    return res;
}
int n;
int main(){
    sca(n);
    rep(i,0,n){
        sca(pre[i]);
        in[i]=pre[i];
    }
    sort(in,in+n);
    if(build(0,n-1,0,n-1,0)){
        puts("YES");
        rep(i,0,n) printf("%d%c",post[i],i==n-1?'\n':' ');
    }
    else{
        reverse(in,in+n);
        cnt=0;
        if(build(0,n-1,0,n-1,1)){
            puts("YES");
            rep(i,0,n) printf("%d%c",post[i],i==n-1?'\n':' ');
        }
        else puts("NO");
    }
    return 0;   
}

版本二，老老实实建树版

#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define scl(x) scanf("%lld",&x)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define prs(x) printf("%s\n",(x))
#define prl(x) printf("%lld\n",x)
#define ll long long
#define PII pair<int,int>
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1e5+5;


vector<int>v,pre,preM,post,postM;
struct node{
    int data;
    node*l,*r;
};

void insert1(node*&root,int data){
    if(root==NULL){
        root=new node;
        root->data=data;
        root->l=root->r=NULL;
        return;
    }
    if(data<root->data) insert1(root->l,data);
    else insert1(root->r,data);
}
void insert2(node*&root,int data){
    if(root==NULL){
        root=new node;
        root->data=data;
        root->l=root->r=NULL;
        return;
    }
    if(data<root->data) insert2(root->r,data);
    else insert2(root->l,data);
}

void preOrder(node *root,vector<int>&vi){
    if(root==NULL)return;
    vi.push_back(root->data);
    preOrder(root->l,vi);
    preOrder(root->r,vi);
}
void postOrder(node *root,vector<int>&vi){
    if(root==NULL)return;
    postOrder(root->l,vi);
    postOrder(root->r,vi);
    vi.push_back(root->data);
}

int main(){
    int n,data;
    node*root=NULL,*rootM=NULL;
    sca(n);
    rep(i,0,n){
        sca(data);
        v.push_back(data);
        insert1(root,data);
        insert2(rootM,data);
    }
    preOrder(root,pre);
    preOrder(rootM,preM);
    postOrder(root,post);
    postOrder(rootM,postM);
   if(v==pre){
        puts("YES");
        rep(i,0,post.size()){
            printf("%d%c",post[i],i==post.size()-1?'\n':' ');
        }
    }
    else if(v==preM){
        puts("YES");
        rep(i,0,postM.size()){
            printf("%d%c",postM[i],i==postM.size()-1?'\n':' ');
        }
    }
    else puts("NO");
    return 0;
}