pat a1043 二叉排序树 镜像树

题目：

https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856

https://www.nowcoder.com/pat/5/submission-detail/64125906

一、题目描述

PAT  a1043 Is It a Binary Search Tree (25分)

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

给出N个正整数来作为一棵二又排序树的结点插入顺序，问：这串序列是否是该二叉排序树的先序序列或是该二叉排序树的镜像树的先序序列。

所谓镜像树是指交换二叉树的所有结点的左右子树而形成的树（也即左子树所有结点数据域大于或等于根结点，而根结点数据域小于右子树所有结点的数据域）。

如果是镜像树，则输出YES，并输出对应的树的后序序列；否则，输出NO。 

二、算法实现 

#include<cstdio> 
#include<vector>
using namespace std;

struct Node
{
	int data;
	Node* left,*right;
};

void insert(Node* &root,int data)
{
	if(root==NULL)
	{
		root=new Node;
		root->data=data;
		root->left=root->right=NULL;
		return;
	}
	if(data<root->data) insert(root->left,data);
	else insert(root->right,data);	
}

void preOrder(Node* root,vector<int> &vi)
{
	if(root==NULL) return;
	vi.push_back(root->data);
	preOrder(root->left,vi);
	preOrder(root->right,vi);
}

void preOrderMirror(Node* root,vector<int> &vi)
{
	if(root==NULL) return;
	vi.push_back(root->data);
	preOrderMirror(root->right,vi);	
	preOrderMirror(root->left,vi);
}

void postOrder(Node* root,vector<int> &vi)
{
	if(root==NULL) return;
	postOrder(root->left,vi);
	postOrder(root->right,vi);
	vi.push_back(root->data);
}

void postOrderMirror(Node* root,vector<int> &vi)
{
	if(root==NULL) return;
	postOrderMirror(root->right,vi);
	postOrderMirror(root->left,vi);
	vi.push_back(root->data);
}


int main()
{
	int n,data;
	vector<int> origin,pre,preM,post,postM;
	Node* root=NULL;  //定义头结点 
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&data);
		origin.push_back(data);
		insert(root,data);
	}
	
	preOrder(root,pre);
	preOrderMirror(root,preM);
	postOrder(root,post);
	postOrderMirror(root,postM);
	if(origin==pre)
	{
		printf("YES\n");
		for(int i=0;i<post.size();i++)
		{
			printf("%d",post[i]);
			if(i<post.size()-1) printf(" ");
		}
	}
	else if(origin==preM)
	{
		printf("YES\n");
		for(int i=0;i<postM.size();i++)
		{
			printf("%d",postM[i]);
			if(i<postM.size()-1) printf(" ");
		}
	}
	else 
	{
		printf("NO\n");
	}
	return 0;
}