二叉树排序查找

本文介绍了一种基于特定序列构建二叉搜索树的方法,并通过该树解决寻找特定房间的问题。利用链表和数组实现算法,帮助精灵邮递员找到正确的路径。

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Elven Postman

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)


Problem Description
Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on trees. However, there is something about them you may not know. Although delivering stuffs through magical teleportation is extremely convenient (much like emails). They still sometimes prefer other more “traditional” methods.

So, as a elven postman, it is crucial to understand how to deliver the mail to the correct room of the tree. The elven tree always branches into no more than two paths upon intersection, either in the east direction or the west. It coincidentally looks awfully like a binary tree we human computer scientist know. Not only that, when numbering the rooms, they always number the room number from the east-most position to the west. For rooms in the east are usually more preferable and more expensive due to they having the privilege to see the sunrise, which matters a lot in elven culture.

Anyways, the elves usually wrote down all the rooms in a sequence at the root of the tree so that the postman may know how to deliver the mail. The sequence is written as follows, it will go straight to visit the east-most room and write down every room it encountered along the way. After the first room is reached, it will then go to the next unvisited east-most room, writing down every unvisited room on the way as well until all rooms are visited.

Your task is to determine how to reach a certain room given the sequence written on the root.

For instance, the sequence 2, 1, 4, 3 would be written on the root of the following tree.

 

Input
First you are given an integer T(T10) indicating the number of test cases.

For each test case, there is a number n(n1000) on a line representing the number of rooms in this tree. n integers representing the sequence written at the root follow, respectively a1,...,an where a1,...,an{1,...,n}.

On the next line, there is a number q representing the number of mails to be sent. After that, there will be q integers x1,...,xq indicating the destination room number of each mail.
 

Output
For each query, output a sequence of move (E or W) the postman needs to make to deliver the mail. For that E means that the postman should move up the eastern branch and W the western one. If the destination is on the root, just output a blank line would suffice.

Note that for simplicity, we assume the postman always starts from the root regardless of the room he had just visited.
 

Sample Input
2 4 2 1 4 3 3 1 2 3 6 6 5 4 3 2 1 1 1
 

Sample Output
E WE
EEEEE
解题报告:
先建立一个排序的二叉树,然后根据建立的结点进行查找路径;
用数组和链表都行,为了练习下链表就用链表写的,写的真艰难,对于链表太不熟悉了,需要多练习啊!
#include<iostream>
#include<cstring>
using namespace std; 
int const maxn=1e3+5;
int num[maxn];
int T,n,q;
struct node{
	int data;
	node *left,*right;
	node(){
		data=0;
		left=NULL,right=NULL;
	}
};
 
void buildtree(node *tree, int num)  
{  
    if(num>tree->data)  
    {  
        if(tree->right==NULL)  
        {  
            tree->right=new node();  
            tree->right->data=num;  
        }  
        else  
            buildtree(tree->right,num);  
    }  
    if(num<tree->data)  
    {  
        if(tree->left==NULL)  
        {  
            tree->left=new node();  
            tree->left->data=num;  
        }  
        else  
            buildtree(tree->left,num);  
    }  
}

void DFS(node *tree,int post)
{
	if(tree==NULL)
	    return;
	if(post>tree->data)
	{
		cout<<"W";
		DFS(tree->right,post);
	}
	else if(post<tree->data)
	{
		cout<<"E";
		DFS(tree->left,post);
	}
	else
	    return;
	
}

int main()
{
	cin>>T;
	while(T--)
	{
		node *root=new node();
		cin>>n;
		for(int j=0;j<n;j++)
		{
		    cin>>num[j];
		}
		root->data=num[0];
		for(int i=1;i<n;i++)
		{
			buildtree(root,num[i]);
		}
		cin>>q;
		while(q--)
		{
			int post;
			cin>>post;
			DFS(root,post);
			cout<<endl;
		}
	}
	return 0;
}




顺序查找、折半查找二叉树排序查找都是常见的查找算法,它们都有各自的优缺点和适用场景。 1. 顺序查找 顺序查找也称为线性查找,是最简单的一种查找方法。它的基本思想是从待查找的数据元素序列中,从头到尾依次逐个比较数据元素的关键字,直到找到目标元素或查找完整个序列。 优点:实现简单,适用于小规模数据或数据元素随机分布的情况。 缺点:时间复杂度为O(n),性能较低,不适用于大规模数据的查找。 2. 折半查找 折半查找也称为二分查找,是一种比较高效的查找算法。它的基本思想是将待查找的数据元素序列按照关键字大小的顺序排列,然后每次取中间位置的元素进行比较,根据比较结果缩小查找范围,直到找到目标元素或查找失败。 优点:时间复杂度为O(log n),性能较高,适用于有序数据的查找。 缺点:要求数据元素必须事先按照关键字大小的顺序排列,同时只适用于静态数据,不能动态地插入或删除元素。 3. 二叉树排序查找 二叉树排序查找也称为二叉排序查找,是一种基于二叉树数据结构的查找算法。它的基本思想是将待查找的数据元素按照关键字大小依次插入到二叉排序树中,然后每次根据目标元素的关键字在二叉排序树上进行搜索,直到找到目标元素或查找失败。 优点:时间复杂度为O(log n),性能较高,同时可以动态地插入、删除元素。 缺点:二叉排序树的平衡性较差,可能会导致树的高度过高,从而影响查找性能。 三种算法的异同和应用: 异同: 顺序查找、折半查找二叉树排序查找都是常见的查找算法,它们的主要区别在于数据的组织方式和查找过程的实现。顺序查找适用于小规模数据或随机分布的数据;折半查找适用于静态有序数据;二叉树排序查找适用于动态的有序数据。 应用: 顺序查找适用于数据量较小的情况,例如在一个长度较短的数组或链表中查找元素;折半查找适用于静态数据的查找,例如在一个已经排好序的数组或链表中查找元素;二叉树排序查找适用于动态数据的查找,例如在一个需要频繁插入、删除元素的数据集合中查找元素。
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