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PAT A1043 Is It a Binary Search Tree (25分)

2020/8/31 14:47:34 文章标签:

PAT A1043 Is It a Binary Search Tree 25分

  • 原题
  • 大体题意
  • 思路
  • 代码
  • 运行结果

原题

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

大体题意

给定一组结点序列,判断是否是二叉搜索树或者其镜像树的前序遍历,如果是要求输出其后序遍历
二叉搜索树的镜像就是所有结点的左右子树互换

思路

无论给定的是不是镜像二叉搜索树的前序序列,只要是正确的序列,根据其建造出来的二叉搜索树都是唯一且正确的。
因此只要将给定的结点序列建树,然后得出该树的前序序列和镜像树的前序序列与给定的结点序列相比较即可。
如果该树的前序序列或镜像树的前序序列与给定的结点序列相同,那么给出该树的后序序列或镜像树的后序序列即可,如果不同则是错误的。
关于镜像树,它的操作也很简单,把树的左右孩子顺序互调即可。
其次,为了快速且方便的将两个序列进行比较,使用vector来保存各种序列。

代码

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

typedef struct node
{
	int data;
	struct node* lchild, *rchild;
}node, *tree;

void create_bst(tree& T, int data)
{
	if (T == NULL)
	{
		T = new node;
		T->data = data;
		T->lchild = T->rchild = NULL;
	}
	else if (data < T->data)
		create_bst(T->lchild, data);
	else
		create_bst(T->rchild, data);
}
void get_pre(tree T,vector<int> &v)
{
	if (T == NULL)
		return;
	v.push_back(T->data);
	get_pre(T->lchild,v);
	get_pre(T->rchild,v);
}
void get_mirrorpre(tree T, vector<int> &v)
{
	if (T == NULL)
		return;
	v.push_back(T->data);
	get_mirrorpre(T->rchild, v);
	get_mirrorpre(T->lchild, v);
}
void get_post(tree T, vector<int>&v)
{
	if (T == NULL)
		return;
	get_post(T->lchild, v);
	get_post(T->rchild, v);
	v.push_back(T->data);
}
void get_mirrorpost(tree T, vector<int>&v)
{
	if (T == NULL)
		return;
	get_mirrorpost(T->rchild, v);
	get_mirrorpost(T->lchild, v);
	v.push_back(T->data);
}
vector<int> yuan, pre, mirror_pre, post, mirror_post;
int main()
{
	int n, a;
	cin >> n;
	tree T = NULL;
	for (int i = 0; i < n; i++)
	{
		cin >> a;
		yuan.push_back(a);
		create_bst(T, a);
	}
	get_pre(T,pre);
	get_mirrorpre(T, mirror_pre);
	if (yuan == pre)
	{
		cout << "YES" << endl;
		get_post(T, post);
		for (int i = 0; i < post.size(); i++)
		{
			if (i != 0)
				cout << " ";
			cout << post[i];
		}
		cout << endl;
	}
	else if (yuan == mirror_pre)
	{
		cout << "YES" << endl;
		get_mirrorpost(T, mirror_post);
		for (int i = 0; i < mirror_post.size(); i++)
		{
			if (i != 0)
				cout << " ";
			cout << mirror_post[i];
		}
		cout << endl;
	}
	else
		cout << "NO" << endl;
	return 0;
}

运行结果

在这里插入图片描述


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