A1043 Is It a Binary Search Tree [bst]

题目思路：先建树，先序遍历一遍和输入的比较判断是yes还是no，然后注意镜像树先序就是先遍历右边的 后序也先遍历右边的

#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#include<math.h>
#include<algorithm>
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 preorder1(node* root, vector<int>& v1)
{
	if (root == NULL) return;
	v1.push_back(root->data);
	preorder1(root->left, v1);
	preorder1(root->right, v1);
}
void preorder2(node* root, vector<int>& v1)
{
	if (root == NULL) return;
	v1.push_back(root->data);
	preorder2(root->right, v1);
	preorder2(root->left, v1);
}
void postorder1(node* root, vector<int>& v1)
{
	if (root == NULL) return;
	postorder1(root->left, v1);
	postorder1(root->right, v1);
	v1.push_back(root->data);
}
void postorder2(node* root, vector<int>& v1)
{
	if (root == NULL) return;
	postorder2(root->right, v1);
	postorder2(root->left, v1);
	v1.push_back(root->data);
}
vector<int>v0,v1, v2, v3, v4;
int main()
{
	int n, data;
	cin >> n;
	node* root = NULL;
	for (int i = 0; i < n; i++)
	{
		cin >> data;
		v0.push_back(data);
		insert(root, data);
	}
	preorder1(root, v1);
	preorder2(root, v2);
	postorder1(root, v3);
	postorder2(root, v4);
	if (v0 == v1)
	{
		cout << "YES" << endl;
		for (int i = 0; i < v3.size(); i++)
		{
			cout << v3[i];
			if (i < v3.size() - 1)
				cout << " ";
		}
	}
	else if (v0 == v2)
	{
		cout << "YES" << endl;
		for (int i = 0; i < v4.size(); i++)
		{
			cout << v4[i];
			if (i < v4.size() - 1)
				cout << " ";
		}
	}
	else
	{
		cout << "NO" << endl;
	}
}