PAT甲级1043

• 1043 Is It a Binary Search Tree (25 分)
• • 题目
  • 输入输出
  • 代码

1043 Is It a Binary Search Tree (25 分)

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.

输入输出

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

代码

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
struct node {
	int data;
	node* lchild;
	node* rchild;
}tnode[1005];
int n;
int value[1005];
int num = 0;
int flag = 0;//判断是否为二叉树
node* create1(int preL, int preR)
{
	if (preL > preR)
		return NULL;
	node* now = new node;
	now->data = value[preL];
	int i;
	for (i = preL + 1; i <= preR; i++)
	{
		if (value[i] >= value[preL])
		{
			break;
		}
	}
	for (int j = i; j <= preR; j++)
	{
		if (value[j] < value[preL])
		{
			flag = 1;
			break;
		}
	}
	i = i - 1;//记录左子树的终点
	now->lchild = create1(preL + 1, i);
	now->rchild = create1(i + 1, preR);
	return now;
}
node* create2(int preL, int preR)
{
	if (preL > preR)
		return NULL;
	node* now = new node;
	now->data = value[preL];
	int i;
	for (i = preL + 1; i <= preR; i++)
	{
		if (value[i] < value[preL])
		{
			break;
		}
	}
	for (int j = i; j <= preR; j++)
	{
		if (value[j] >= value[preL])
		{
			flag = 1;
			break;
		}
	}
	i = i - 1;//记录左子树的终点
	now->lchild = create2(preL + 1, i);
	now->rchild = create2(i + 1, preR);
	return now;
}
void postorder(node* root)
{
	if (root == NULL)
		return;
	if(root->lchild!=NULL)
		postorder(root->lchild);
	if (root->rchild != NULL)
		postorder(root->rchild);
	cout << root->data;

	num++;
	if (num < n)
		cout << " ";
}
int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> value[i];
	node* root;
	if(value[n-1]>value[0])
		root=create1(0, n-1);
	else root = create2(0, n - 1);
	if (flag == 0)
	{
		cout << "YES"<<endl;
		postorder(root);
	}
	else cout << "NO";
}