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编程算法 - 二叉搜索树(binary search tree) 代码(C)

2015-10-09 21:39 369 查看

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\n", isOK?

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

return 0;
}

输出;

result = Yes
result = No
result = Yes


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