1043. Is It a Binary Search Tree (25)-优快云博客

本文介绍了一个程序，用于判断给定的整数序列是否为二叉搜索树的前序遍历或其镜像的前序遍历，并输出相应的后序遍历序列。通过递归检查每个节点的左右子树，确保它们符合二叉搜索树的定义。

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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

思路：

这道题我没有创建二叉树。因为题目要求判断序列是否是前序还是镜像前序，如果是输出其后序排列。程序里在判断其是否为前序或是镜像前序时，就把各节点按照后序顺序输入。

代码：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
//利用左子树的点均小于节点值，右子树的点均大于等于节点的值来判断

vector <int> q(1000);
vector<int> q_post;
bool cmp_pre(const int &n1, const int &n2)
{
	return n1 < n2;
}

bool cmp_mirr(const int &n1, const int &n2)
{
	return n1 >= n2;
}

//判断是否满足条件
bool issatified = true;
bool is_satified(int root, int end, bool cmp(const int &, const int &))
{

	if (!issatified || root>end)
	{
		return 0;
	}
	else
	{
		if (root == end || (root + 1 == end))
		{
			if (root != end)
				q_post.push_back(q[end]);
			return 1;
		}
		else
		{
			int i, r;
			//找到右子树节点
			for (i = root + 1; i <= end; ++i)
			{
				if (!cmp(q[i], q[root]))
					break;
			}
			//判断左子树
			if (i != root + 1)   //有左子树
			{
				is_satified(root + 1, i - 1, cmp);
				q_post.push_back(q[root + 1]);//插入左子树节点
			}
			//在，左子树满足条件的同时，有右子树，判断右子树是否满足条件
			if (issatified && i <= end)
			{
				r = i;//右子树节点
				if (r != end)
				{
					for (i = r + 1; i <= end; ++i)
					{
						if (cmp(q[i], q[root]))
						{
							issatified = 0;
							return 0;
						}
					}
				}
				if (issatified)
				{
					is_satified(r, end, cmp);
					q_post.push_back(q[r]);//插入右子树节点
				}
			}

		}
	}
}

int main()
{
	int N;
	cin >> N;
	q.resize(N);
	int i = 0;
	for (; i < N; ++i)
	{
		cin >> q[i];
	}

	//判断是否为前序遍历
	q_post.clear();
	issatified = true;
	is_satified(0, N - 1, cmp_pre);
	if (!issatified)
	{
		q_post.clear();
		issatified = true;
		is_satified(0, N - 1, cmp_mirr);
		if (!issatified)
		{
			cout << "NO" << endl;
		}
	}
	if (issatified)
	{
		cout << "YES" << endl;
		q_post.push_back(q[0]);
		for (i = 0; i < q_post.size(); ++i)
		{
			cout << q_post[i];
			if (i == q_post.size() - 1)
				cout << endl;
			else
				cout << " ";
		}
	}

	system("pause");
	return 0;
}