编程算法

二叉搜索树(binary search tree) 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

二叉搜索树(binary search tree)能够高效的进行插入, 查询, 删除某个元素, 时间复杂度O(logn).

简单的实现方法例如以下.

代码:

/*
 * main.cpp
 *
 *  Created on: 2014.7.20
 *      Author: spike
 */

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>

#include <queue>
#include <vector>
#include <functional>

using namespace std;

struct node {
	int val;
	node *lch, *rch;
};

class BinarySearchTree {
public:
	node *insert(node *p, int x) {
		if (p==NULL) {
			node* q = new node;
			q->val = x;
			q->lch = q->rch = NULL;
			return q;
		} else {
			if (x<p->val) p->lch = insert(p->lch, x);
			else p->rch = insert(p->rch, x);
			return p;
		}
	}
	bool find(node *p, int x) {
		if (p==NULL) return false;
		else if (x==p->val) return true;
		else if (x<p->val) return find(p->lch, x);
		else return find(p->rch, x);
	}
	node *remove(node *p, int x) {
		if (p==NULL) return NULL;
		else if (x<p->val) p->lch = remove(p->lch, x);
		else if (x>p->val) p->rch = remove(p->rch, x);
		else if (p->lch == NULL) {
			node* q=p->rch;
			delete p;
			return q;
		} else if (p->lch->rch == NULL) {
			node* q = p->lch;
			q->rch = p->rch;
			delete p;
			return q;
		} else {
			node* q;
			for (q = p->lch; q->rch->rch!=NULL; q=q->rch);
			node* r = q->rch;
			q->rch = r->lch;
			r->lch = p->lch;
			r->rch = p->rch;
			delete p;
			return r;
		}
		return p;
	}
};


int main(void)
{
	BinarySearchTree iBST;
	node* root = NULL;
	root = iBST.insert(root, 7);
	root = iBST.insert(root, 2);
	root = iBST.insert(root, 15);
	root = iBST.insert(root, 1);
	root = iBST.insert(root, 5);
	root = iBST.insert(root, 10);
	root = iBST.insert(root, 17);
	root = iBST.insert(root, 4);
	root = iBST.insert(root, 6);
	root = iBST.insert(root, 8);
	root = iBST.insert(root, 11);
	root = iBST.insert(root, 16);
	root = iBST.insert(root, 19);

	bool isOK = iBST.find(root, 15);
	printf("result = %s
", isOK?
"Yes":"No");
	iBST.remove(root,15);
	isOK = iBST.find(root, 15);
	printf("result = %s
", isOK?"Yes":"No");
	isOK = iBST.find(root, 10);
	printf("result = %s
", isOK?"Yes":"No");

	return 0;
}

输出;

result = Yes
result = No
result = Yes

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