点击链接PAT甲级-AC全解汇总
题目：
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
- The left subtree of a node contains only nodes with keys less than the node’s key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
- Both the left and right subtrees must also be binary search trees.
If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.
Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.
Output Specification:
For each test case, first print in a line YES if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or NO if not. Then if the answer is YES, print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input 1:
7
8 6 5 7 10 8 11
Sample Output 1:
YES
5 7 6 8 11 10 8
Sample Input 2:
7
8 10 11 8 6 7 5
Sample Output 2:
YES
11 8 10 7 5 6 8
Sample Input 3:
7
8 6 8 5 10 9 11
Sample Output 3:
NO
题意：
输入一个数，组成查找树，并判断输入的顺序是先序或者镜像先序，如果是的话，输出后序或镜像后序，如果不是直接输出NO
让我自己写可能写不到算法笔记那么好，我直接附上之前算法笔记的代码好了。
算法笔记的代码：
#include<bits/stdc++.h>
using namespace std;
#define null 0
struct node{
int data;node *lchild,*rchild;
};
vector<int>origin,pre,prem,post,postm;
void my_p(vector<int>vec){
for(auto it:vec)cout<<it<<" ";cout<<endl;
}
void insert(node* &root,int data){
if(root==null){
root=new node;root->data=data;root->lchild=root->rchild=null;return;}if(data<root->data)insert(root->lchild,data);elseinsert(root->rchild,data);
}//preorder
void preorder(node* root,vector<int>&vi){
if(root==null)return;vi.push_back(root->data);preorder(root->lchild,vi);preorder(root->rchild,vi);
}
//preordernmirror
void preordernmirror(node* root,vector<int>&vi){
if(root==null)return;vi.push_back(root->data);preordernmirror(root->rchild,vi);preordernmirror(root->lchild,vi);
}
//postorder
void postorder(node* root,vector<int>&vi){
if(root==null)return;postorder(root->lchild,vi);postorder(root->rchild,vi);vi.push_back(root->data);
}
//postordernmirror
void postordernmirror(node* root,vector<int>&vi){
if(root==null)return;postordernmirror(root->rchild,vi);postordernmirror(root->lchild,vi);vi.push_back(root->data);
}int main(){
int n,data;node* root=null;scanf("%d",&n);for(int i=0;i<n;i++){
scanf("%d",&data);origin.push_back(data);insert(root,data);}preorder(root,pre);preordernmirror(root,prem);postorder(root,post);postordernmirror(root,postm);if(origin==pre){
printf("YES\n");for(int i=0;i<post.size();i++){
if(i>0)printf(" ");printf("%d",post[i]);}}else if(origin==prem){
printf("YES\n");for(int i=0;i<postm.size();i++){
if(i>0)printf(" ");printf("%d",postm[i]);}}else printf("NO\n");return 0;
}