04-树4. Search in a Binary Search Tree (25)

<span style="background-color: rgb(250, 250, 250); font-family: 'Droid Sans', Verdana, 'Microsoft YaHei', Tahoma, sans-serif; line-height: 1em;"><span style="font-size:24px;">04-树4. Search in a Binary Search Tree (25)</span></span>

时间限制

100 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

To search a key in a binary search tree, we start from the root and move all the way down, choosing branches according to the comparison results of the keys. The searching path corresponds to a sequence of keys. For example, following {1, 4, 2, 3} we can find 3 from a binary search tree with 1 as its root. But {2, 4, 1, 3} is not such a path since 1 is in the right subtree of the root 2, which breaks the rule for a binary search tree. Now given a sequence of keys, you are supposed to tell whether or not it indeed correspnds to a searching path in a binary search tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N and M (<=100) which are the total number of sequences, and the size of each sequence, respectively. Then N lines follow, each gives a sequence of keys. It is assumed that the keys are numbered from 1 to M.

Output Specification:

For each sequence, print in a line "YES" if the sequence does correspnd to a searching path in a binary search tree, or "NO" if not.

Sample Input:

3 4
1 4 2 3
2 4 1 3
3 2 4 1

Sample Output:

YES
NO
NO

思路：题目的意思是判断是否是一个二叉搜索树序列，首先我想到的是，建立一个二维数组存储数据，然后用两个循环判断输入的每一个key。显然这个方法并不好。既然已经学会了二叉树的建立，那就用递归的方法实现实时判断。PAT的测试也并不要求一定要将输入全部完成才能输出。明确了之后只需根据二叉搜索树的特点，在每次插入的时候判断其另一子树是否为空就好。

代码：

#include<iostream>
using namespace std;

typedef int ElementType;

typedef struct TreeNode *BinTree;
typedef BinTree Position;
typedef struct TreeNode{
	ElementType data;
	BinTree Lchild;
	BinTree Rchild;
};

bool Insert_BST(BinTree *T, int item);

int main(int argc, char *argv[])
{
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < N; i++)
	{
		bool flag = true;
		BinTree T = NULL;

		for (int j = 0; j < M; j++)
		{
			int n;
			cin >> n;
			if (flag&&!Insert_BST(&T, n))
				flag = false;
		}
		if (flag)
			cout << "YES" << endl;
		else
			cout << "NO" << endl;
	}

	system("pause");

	return 0;
}

bool Insert_BST(BinTree *T, int item)
{
	if (!*T)
	{
		*T = (BinTree)malloc(sizeof(struct TreeNode));
		(*T)->data = item;
		(*T)->Lchild = (*T)->Rchild = NULL;
	}
	else
	{//Judge
		if (item < (*T)->data && (*T)->Rchild == NULL)
		{
			if (!Insert_BST(&(*T)->Lchild, item))
				return false;
		}
		else if (item>(*T)->data && (*T)->Lchild == NULL)
		{
			if (!Insert_BST(&(*T)->Rchild, item))
				return false;
		}
		else
			return false;//wrong,dose not insert
	}

	return true;
}