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法律研究C++ 程式來實現 B 樹\n
B 樹是二叉搜尋樹的推廣，其中一個節點可以有多於兩個子節點。它基本上是一個自我平衡樹資料結構，可以維護已排序資料，並允許在對數時間內進行順序訪問、搜尋、插入和刪除。
這是一個 C++ 程式，用於實現 6 階 B 樹。
演算法
Begin function insert() to insert the nodes into the tree: Initialize x as root. if x is leaf and having space for one more info then insert a to x. else if x is not leaf, do Find the child of x that is going to be traversed next. If the child is not full, change x to point to the child. If the child is full, split it and change x to point to one of the two parts of the child. If a is smaller than mid key in the child, then set x as first part of the child. Else second part of the child. When split the child, move a key from the child to its parent x. End
示例程式碼
#include<iostream>
using namespace std;
struct BTree//node declaration {
int *d;
BTree **child_ptr;
bool l;
int n;
}*r = NULL, *np = NULL, *x = NULL;
BTree* init()//creation of node {
int i;
np = new BTree;
np->d = new int[6];//order 6
np->child_ptr = new BTree *[7];
np->l = true;
np->n = 0;
for (i = 0; i < 7; i++) {
np->child_ptr[i] = NULL;
}
return np;
}
void traverse(BTree *p)//traverse the tree {
cout<<endl;
int i;
for (i = 0; i < p->n; i++) {
if (p->l == false) {
traverse(p->child_ptr[i]);
}
cout << " " << p->d[i];
}
if (p->l == false) {
traverse(p->child_ptr[i]);
}
cout<<endl;
}
void sort(int *p, int n)//sort the tree {
int i, j, t;
for (i = 0; i < n; i++) {
for (j = i; j <= n; j++) {
if (p[i] >p[j]) {
t = p[i];
p[i] = p[j];
p[j] = t;
}
}
}
}
int split_child(BTree *x, int i) {
int j, mid;
BTree *np1, *np3, *y;
np3 = init();//create new node
np3->l = true;
if (i == -1) {
mid = x->d[2];//find mid
x->d[2] = 0;
x->n--;
np1 = init();
np1->l= false;
x->l= true;
for (j = 3; j < 6; j++) {
np3->d[j - 3] = x->d[j];
np3->child_ptr[j - 3] = x->child_ptr[j];
np3->n++;
x->d[j] = 0;
x->n--;
}
for (j = 0; j < 6; j++) {
x->child_ptr[j] = NULL;
}
np1->d[0] = mid;
np1->child_ptr[np1->n] = x;
np1->child_ptr[np1->n + 1] = np3;
np1->n++;
r = np1;
} else {
y = x->child_ptr[i];
mid = y->d[2];
y->d[2] = 0;
y->n--;
for (j = 3; j <6 ; j++) {
np3->d[j - 3] = y->d[j];
np3->n++;
y->d[j] = 0;
y->n--;
}
x->child_ptr[i + 1] = y;
x->child_ptr[i + 1] = np3;
}
return mid;
}
void insert(int a) {
int i, t;
x = r;
if (x == NULL) {
r = init();
x = r;
} else {
if (x->l== true && x->n == 6) {
t = split_child(x, -1);
x = r;
for (i = 0; i < (x->n); i++) {
if ((a >x->d[i]) && (a < x->d[i + 1])) {
i++;
break;
} else if (a < x->d[0]) {
break;
} else {
continue;
}
}
x = x->child_ptr[i];
} else {
while (x->l == false) {
for (i = 0; i < (x->n); i++) {
if ((a >x->d[i]) && (a < x->d[i + 1])) {
i++;
break;
} else if (a < x->d[0]) {
break;
} else {
continue;
}
}
if ((x->child_ptr[i])->n == 6) {
t = split_child(x, i);
x->d[x->n] = t;
x->n++;
continue;
} else {
x = x->child_ptr[i];
}
}
}
}
x->d[x->n] = a;
sort(x->d, x->n);
x->n++;
}
int main() {
int i, n, t;
cout<<"enter the no of elements to be inserted\n";
cin>>n;
for(i = 0; i < n; i++) {
cout<<"enter the element\n";
cin>>t;
insert(t);
}
cout<<"traversal of constructed B tree\n";
traverse(r);
}輸出
enter the no of elements to be inserted 7 enter the element 10 enter the element 20 enter the element 30 enter the element 40 enter the element 50 enter the element 60 enter the element 70 traversal of constructed B tree 10 20 30 40 50 60 70
更新日期： 2019 年 7 月 30 日
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