B树的定义:
1、根节点至少有两个分支
2、除了根节点以外,所有节点的关键字个数至少为M/2个,最多为M-1
3、每个节点的度数均是关键字数加一
4、所有的叶子节点都在同一层
插入:
我们设计节点的结构如下:
#define M 5#define MAX M - 1#define MIN M/2typedef char KeyType;typedef struct {}Record;typedef struct ElemType{ KeyType key; Record *recptr;}ElemType;typedef struct BNode{ int num; BNode *parent; ElemType data[M+1]; BNode*sub[M+1];}BNode,*BTree;typedef struct Result{ bool tag; BNode*pnode; int index;}Result;
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插入代码:
当以个节点的个数大于MAX时就分裂,如果根节点分裂会产生新根,否则就将分裂出来的节点插入到双亲中,如果双亲又大于MAX就继续分裂,这样就能保证B树的定义的正确性,代码如下:
ElemType MoveItem(BNode*ptr, BNode *s, int pos){ int tmp = ptr->num; for (int i = pos + 1, j = 0; i <= tmp; i++, j++) { s->data[j] = ptr->data[i]; s->sub[j] = ptr->sub[i]; if (ptr->sub[i] != NULL) { s->sub[j]->parent = s; } } s->parent = ptr->parent; s->num = ptr->num = MIN; return s->data[0];}BNode * MakeRoot(ElemType x, BNode *left, BNode *right){ BNode *s = BuyNode(); s->num = 1; s->data[1] = x; s->sub[0] = left; if (left != NULL) left->parent = s; s->sub[1] = right; if (right != NULL) right->parent = s; return s;}bool InsertItem(BNode*ptr, int pos, ElemType e, BNode*right);BNode *Splice(BNode*ptr){ BNode *s = BuyNode(); ElemType e = MoveItem(ptr, s, MIN); if (ptr->parent == NULL) { return MakeRoot(e,ptr, s); } else { ptr = ptr->parent; int i = ptr->num; ptr->data[0] = e;//这句很关键,如果ptr->data[0]未设置就会和0位置比较还没有结果,插入位置就会出错 while (ptr->data[i].key > s->data[0].key) --i; InsertItem(ptr, i + 1, s->data[0], s); if (ptr->num > MAX) { return Splice(ptr); } return NULL; }}bool InsertItem(BNode*ptr, int pos, ElemType e, BNode*right)//BNode&node{ for (int i = ptr->num; i >= pos; --i) { ptr->data[i + 1] = ptr->data[i]; ptr->sub[i + 1] = ptr->sub[i]; } // ptr->data[pos] = e; ptr->sub[pos] = right; if (right != NULL) { right->parent = ptr; }// ptr->num += 1; return true;}bool Insert(BTree *ptr, ElemType e){ if (ptr == NULL) return false; if (*ptr == NULL) { *ptr = MakeRoot(e, NULL, NULL); return true; } Result res=FindValue(*ptr, e.key); if (res.pnode == NULL || res.tag) return false; InsertItem(res.pnode, res.index+1, e, NULL); if (res.pnode->num > MAX) { BNode*p = Splice(res.pnode); if (p != NULL) { *ptr = p; } } return true;}
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辅助函数:
BNode* BuyNode(){ BNode *node = new BNode(); if (node == NULL) exit(-1); memset(node, 0, sizeof(BNode)); return node;}Result FindValue(BNode*ptr, KeyType e){ Result res = { false, NULL, -1 }; while (ptr != NULL) { int i = ptr->num; ptr->data[0].key = e; while (ptr->data[i].key > e) --i; res.pnode = ptr; res.index = i; if (i != 0 && ptr->data[i].key == e) { res.tag = true; break; } else ptr = ptr->sub[i]; } return res;}
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B树的删除
B树的删除,我们将带有分支的节点中的关键码删除,用他的前驱和后继替换掉这个被删除的关键码,然后删除前驱或者后继,删除前驱或者后继之后,会出现与B树定义不相符的情况,比如关键码个数小于MIN的情况,这个时候就要做相应的旋转,如过旋转不了就只有进行节点的合并,合并有可能会产生新根,代码如下:
//找前驱BNode *FindPre(BNode*ptr){ while (ptr!=NULL&&ptr->sub[ptr->num] != NULL) { ptr = ptr->sub[ptr->num]; } return ptr;}//找后继BNode *FindNext(BNode*ptr){ while (ptr != NULL&&ptr->sub[0] != NULL) { ptr = ptr->sub[0]; } return ptr;}//删除叶子结点void DelLeafItem(BNode *ptr, int pos){ for (int i = pos; i < ptr->num; i++) { ptr->data[i] = ptr->data[i + 1]; ptr->sub[i] = ptr->sub[i + 1]; } ptr->num -= 1;}//右旋转void RightRotateLeaf(BNode *leftbro, BNode*ptr, BNode *parent, int pos){ ptr->data[0] = parent->data[pos]; for (int i = ptr->num; i >= 0; i--) { ptr->data[i + 1] = ptr->data[i]; ptr->sub[i + 1] = ptr->sub[i]; } ptr->num += 1; ptr->sub[0] = leftbro->sub[leftbro->num]; if (ptr->sub[0] != NULL)// { ptr->sub[0]->parent = ptr; } parent->data[pos] = leftbro->data[leftbro->num]; leftbro->num -= 1;}//左旋转void LeftRotateLeaf(BNode *rightbro,BNode *ptr,BNode *parent,int pos){ ptr ->data[ptr->num+1] = parent->data[pos + 1]; ptr->sub[ptr->num + 1] = rightbro->sub[0]; if (ptr->sub[ptr->num+1]!=NULL) { ptr->sub[ptr->num + 1]->parent = ptr; } ptr->num += 1; parent->data[pos + 1] = rightbro->data[1]; for (int i =0; i < rightbro->num; i++) { rightbro->data[i] = rightbro->data[i + 1]; rightbro->sub[i] = rightbro->sub[i + 1]; } rightbro->num -= 1;}//向左合并void LeftMerge(BNode*leftbro, BNode*ptr, BNode*parent, int pos){ ptr->data[0] = parent->data[pos]; for (int i = 0,j=leftbro->num+1; i <= ptr->num; i++,j++) { leftbro->data[j] = ptr->data[i]; leftbro->sub[j] = ptr->sub[i]; if (leftbro->sub[j] != NULL) { leftbro->sub[j]->parent = leftbro; } } leftbro->num = leftbro->num + ptr->num + 1; free(ptr); DelLeafItem(parent, pos);}//向右合并void RightMerge(BNode *ptr, BNode *rightbro, BNode *parent, int pos){ LeftMerge(ptr, rightbro, parent, pos+1);}//出现小于MIN的情况的调整函数BNode *AdjusLeaf(BNode*ptr){ BNode*parent = ptr->parent; int pos = 0; while (parent->sub[pos] != ptr) ++pos; BNode*leftbro = pos-1<0?NULL:parent->sub[pos-1]; BNode*rightbro = pos+1>=MAX?NULL:parent->sub[pos+1]; if (leftbro!=NULL&&leftbro->num>MIN) { RightRotateLeaf(leftbro,ptr,parent,pos); } else if (rightbro!=NULL&&rightbro->num>MIN) { LeftRotateLeaf(rightbro, ptr,parent, pos); } else if(leftbro!=NULL) { LeftMerge(leftbro, ptr, parent, pos); ptr = leftbro; } else if (rightbro != NULL) { RightMerge(ptr, rightbro, parent, pos); // ptr = rightbro; } if (parent->parent != NULL&&parent->num < MIN) { return AdjusLeaf(parent); } if (parent->parent == NULL&&parent->num <= 0) { free(parent); ptr->parent = NULL; return ptr; } return NULL;}//删除函数void ReMove(BNode*&root, KeyType e){ if (root == NULL) return; Result res = FindValue(root, e); if (res.pnode == NULL || res.tag==false) return; BNode *ptr = res.pnode; int pos = res.index; BNode*pre = FindPre(ptr->sub[pos-1]); BNode*next = FindNext(ptr->sub[pos]); if (pre != NULL&&pre->num > MIN) { ptr->data[pos] = pre->data[pre->num]; ptr = pre; pos = pre->num; } else if (next != NULL&&next->num > MIN) { ptr->data[pos] = next->data[1]; ptr = next; pos = 1; } else if (pre != NULL) { ptr->data[pos] = pre->data[pre->num]; ptr = pre; pos = pre->num; } else if (next != NULL) { ptr->data[pos] = next->data[1]; ptr = next; pos = 1; } DelLeafItem(ptr, pos);// if (ptr->parent != NULL&&ptr->num < MIN) { BNode*newroot = AdjusLeaf(ptr); if (newroot != NULL) { root = newroot; } } else if (ptr->parent == NULL&&ptr->num <= 0) { free(root); root = NULL; }}
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B树的遍历
利用递归的特性 ,代码十分简洁,先递归到最左边,然后打印一个关键码就遍历一个分支,代码如下:
void InOder(BNode*root){ if (root != NULL) { InOder(root->sub[0]); for (int i = 1; i <= root->num; i++) { cout << root->data[i].key; InOder(root->sub[i]); } }}