Dzēšana no B koka

Šajā apmācībā jūs uzzināsiet, kā izdzēst atslēgu no b-koka. Jūs atradīsit arī piemērus, kā izdzēst atslēgas no B koka C, C ++, Java un Python.

Elementa dzēšana B kokā sastāv no trim galvenajiem notikumiem: mezgla meklēšana, kur pastāv dzēšamā atslēga, atslēgas dzēšana un, ja nepieciešams, koka līdzsvarošana.

Dzēšot koku, var rasties stāvoklis, ko sauc par nepietiekamu plūsmu . Nepietiekama plūsma rodas, ja mezglā ir mazāks par minimālo tajā esošo atslēgu skaitu.

Pirms dzēšanas darbības izpētes jāsaprot šādi termini:

  1. Inorder priekštecis
    Lielāko atslēgu mezgla kreisajā bērnā sauc par tā pasūtījuma priekšgājēju.
  2. Pasūtījuma pēctecis
    Mazāko atslēgu mezgla labajā bērniņā sauc par tā pasūtījuma pēcteci.

Dzēšanas darbība

Pirms iziet tālāk norādītās darbības, jāzina šie fakti par B pakāpes m koku .

  1. Mezglā var būt ne vairāk kā m bērni. (ti, 3)
  2. Mezglā var būt ne vairāk kā m - 1atslēgas. (ti, 2)
  3. Mezglā jābūt vismaz ⌈m/2⌉bērniem. (ti, 2)
  4. Mezglā (izņemot saknes mezglu) jābūt vismaz ⌈m/2⌉ - 1atslēgām. (ti, 1)

B kokā ir trīs galvenie dzēšanas gadījumi.

I gadījums

Dzēšamā atslēga atrodas lapā. Tam ir divi gadījumi.

  1. Atslēgas dzēšana nepārkāpj minimālā atslēgu skaita, kas mezglā būtu, īpašību.
    Zemāk esošajā kokā 32 dzēšana nepārkāpj iepriekš minētās īpašības. Lapu atslēgas (32) dzēšana no B koka
  2. Dzēšot atslēgu, tiek pārkāpts minimālā atslēgu skaita rekvizīts, kas mezglā būtu jāglabā. Šajā gadījumā mēs aizņemamies atslēgu no tās tuvākā kaimiņa brāļa vai māsas mezgla secībā no kreisās uz labo.
    Vispirms apmeklējiet tuvāko kreiso brāli. Ja kreisajā brāļa / māsas mezglā ir vairāk nekā minimālais atslēgu skaits, tad aizņemieties atslēgu no šī mezgla.
    Pārējais, pārbaudiet, vai aizņemties no labā brāļa / māsas mezgla.
    Zemāk esošajā kokā, izdzēšot 31, tiek iegūts iepriekš minētais nosacījums. Aizņemsimies atslēgu no kreisā brāļa / māsas mezgla. Lapas atslēgas dzēšana (31) Ja abiem tiešajiem brāļu un māsu mezgliem jau ir minimālais atslēgu skaits, tad apvienojiet mezglu ar kreiso vai labo brāļa vai māsas mezglu. Šī apvienošana tiek veikta caur vecāku mezglu.
    Dzēšot 30 rezultātus, iepriekš minētajā gadījumā.
    Dzēst lapu atslēgu (30)

II gadījums

Ja dzēšamā atslēga atrodas iekšējā mezglā, rodas šādi gadījumi.

  1. Iekšējais mezgls, kas tiek izdzēsts, tiek aizstāts ar pasūtījuma priekšgājēju, ja kreisajam bērnam ir vairāk nekā minimālais atslēgu skaits. Iekšējā mezgla dzēšana (33)
  2. Iekšējais mezgls, kas tiek izdzēsts, tiek aizstāts ar pasūtījuma pēcteci, ja pareizajam bērnam ir vairāk nekā minimālais atslēgu skaits.
  3. Ja kādam bērnam ir tieši minimālais atslēgu skaits, sapludiniet kreiso un labo bērnu.
    Iekšējā mezgla dzēšana (30) Pēc apvienošanas, ja vecāka mezglā ir mazāks par minimālo atslēgu skaitu, meklējiet brāļus un māsas, kā tas ir I gadījumā.

III gadījums

Šajā gadījumā koka augstums samazinās. Ja mērķa atslēga atrodas iekšējā mezglā un atslēgas dzēšana noved pie mazāka atslēgu skaita mezglā (ti, mazāka par nepieciešamo minimumu), tad meklējiet pasūtījuma priekšgājēju un pēcteci. Ja abos bērnos ir minimālais atslēgu skaits, aizņemšanās nevar notikt. Tas noved pie II gadījuma (3), ti, bērnu apvienošanas.

Atkal meklējiet brāli, lai aizņemtos atslēgu. Bet, ja brālim vai māsai ir tikai minimālais atslēgu skaits, apvienojiet mezglu ar brāli un māsu kopā ar vecākiem. Attiecīgi sakārtojiet bērnus (palielinot kārtību).

Iekšējā mezgla dzēšana (10)

Python, Java un C / C ++ piemēri

Python Java C C ++ 
 # Deleting a key on a B-tree in Python # Btree node class BTreeNode: def __init__(self, leaf=False): self.leaf = leaf self.keys = () self.child = () class BTree: def __init__(self, t): self.root = BTreeNode(True) self.t = t # Insert a key def insert(self, k): root = self.root if len(root.keys) == (2 * self.t) - 1: temp = BTreeNode() self.root = temp temp.child.insert(0, root) self.split_child(temp, 0) self.insert_non_full(temp, k) else: self.insert_non_full(root, k) # Insert non full def insert_non_full(self, x, k): i = len(x.keys) - 1 if x.leaf: x.keys.append((None, None)) while i>= 0 and k(0)  = 0 and k(0)  x.keys(i)(0): i += 1 self.insert_non_full(x.child(i), k) # Split the child def split_child(self, x, i): t = self.t y = x.child(i) z = BTreeNode(y.leaf) x.child.insert(i + 1, z) x.keys.insert(i, y.keys(t - 1)) z.keys = y.keys(t: (2 * t) - 1) y.keys = y.keys(0: t - 1) if not y.leaf: z.child = y.child(t: 2 * t) y.child = y.child(0: t - 1) # Delete a node def delete(self, x, k): t = self.t i = 0 while i x.keys(i)(0): i += 1 if x.leaf: if i < len(x.keys) and x.keys(i)(0) == k(0): x.keys.pop(i) return return if i = t: self.delete(x.child(i), k) else: if i != 0 and i + 2 = t: self.delete_sibling(x, i, i - 1) elif len(x.child(i + 1).keys)>= t: self.delete_sibling(x, i, i + 1) else: self.delete_merge(x, i, i + 1) elif i == 0: if len(x.child(i + 1).keys)>= t: self.delete_sibling(x, i, i + 1) else: self.delete_merge(x, i, i + 1) elif i + 1 == len(x.child): if len(x.child(i - 1).keys)>= t: self.delete_sibling(x, i, i - 1) else: self.delete_merge(x, i, i - 1) self.delete(x.child(i), k) # Delete internal node def delete_internal_node(self, x, k, i): t = self.t if x.leaf: if x.keys(i)(0) == k(0): x.keys.pop(i) return return if len(x.child(i).keys)>= t: x.keys(i) = self.delete_predecessor(x.child(i)) return elif len(x.child(i + 1).keys)>= t: x.keys(i) = self.delete_successor(x.child(i + 1)) return else: self.delete_merge(x, i, i + 1) self.delete_internal_node(x.child(i), k, self.t - 1) # Delete the predecessor def delete_predecessor(self, x): if x.leaf: return x.pop() n = len(x.keys) - 1 if len(x.child(n).keys)>= self.t: self.delete_sibling(x, n + 1, n) else: self.delete_merge(x, n, n + 1) self.delete_predecessor(x.child(n)) # Delete the successor def delete_successor(self, x): if x.leaf: return x.keys.pop(0) if len(x.child(1).keys)>= self.t: self.delete_sibling(x, 0, 1) else: self.delete_merge(x, 0, 1) self.delete_successor(x.child(0)) # Delete resolution def delete_merge(self, x, i, j): cnode = x.child(i) if j> i: rsnode = x.child(j) cnode.keys.append(x.keys(i)) for k in range(len(rsnode.keys)): cnode.keys.append(rsnode.keys(k)) if len(rsnode.child)> 0: cnode.child.append(rsnode.child(k)) if len(rsnode.child)> 0: cnode.child.append(rsnode.child.pop()) new = cnode x.keys.pop(i) x.child.pop(j) else: lsnode = x.child(j) lsnode.keys.append(x.keys(j)) for i in range(len(cnode.keys)): lsnode.keys.append(cnode.keys(i)) if len(lsnode.child)> 0: lsnode.child.append(cnode.child(i)) if len(lsnode.child)> 0: lsnode.child.append(cnode.child.pop()) new = lsnode x.keys.pop(j) x.child.pop(i) if x == self.root and len(x.keys) == 0: self.root = new # Delete the sibling def delete_sibling(self, x, i, j): cnode = x.child(i) if i 0: cnode.child.append(rsnode.child(0)) rsnode.child.pop(0) rsnode.keys.pop(0) else: lsnode = x.child(j) cnode.keys.insert(0, x.keys(i - 1)) x.keys(i - 1) = lsnode.keys.pop() if len(lsnode.child)> 0: cnode.child.insert(0, lsnode.child.pop()) # Print the tree def print_tree(self, x, l=0): print("Level ", l, " ", len(x.keys), end=":") for i in x.keys: print(i, end=" ") print() l += 1 if len(x.child)> 0: for i in x.child: self.print_tree(i, l) B = BTree(3) for i in range(10): B.insert((i, 2 * i)) B.print_tree(B.root) B.delete(B.root, (8,)) print("") B.print_tree(B.root)  
 // Inserting a key on a B-tree in Java import java.util.Stack; public class BTree ( private int T; public class Node ( int n; int key() = new int(2 * T - 1); Node child() = new Node(2 * T); boolean leaf = true; public int Find(int k) ( for (int i = 0; i < this.n; i++) ( if (this.key(i) == k) ( return i; ) ) return -1; ); ) public BTree(int t) ( T = t; root = new Node(); root.n = 0; root.leaf = true; ) private Node root; // Search the key private Node Search(Node x, int key) ( int i = 0; if (x == null) return x; for (i = 0; i < x.n; i++) ( if (key < x.key(i)) ( break; ) if (key == x.key(i)) ( return x; ) ) if (x.leaf) ( return null; ) else ( return Search(x.child(i), key); ) ) // Split function private void Split(Node x, int pos, Node y) ( Node z = new Node(); z.leaf = y.leaf; z.n = T - 1; for (int j = 0; j < T - 1; j++) ( z.key(j) = y.key(j + T); ) if (!y.leaf) ( for (int j = 0; j = pos + 1; j--) ( x.child(j + 1) = x.child(j); ) x.child(pos + 1) = z; for (int j = x.n - 1; j>= pos; j--) ( x.key(j + 1) = x.key(j); ) x.key(pos) = y.key(T - 1); x.n = x.n + 1; ) // Insert the key public void Insert(final int key) ( Node r = root; if (r.n == 2 * T - 1) ( Node s = new Node(); root = s; s.leaf = false; s.n = 0; s.child(0) = r; Split(s, 0, r); _Insert(s, key); ) else ( _Insert(r, key); ) ) // Insert the node final private void _Insert(Node x, int k) ( if (x.leaf) ( int i = 0; for (i = x.n - 1; i>= 0 && k  = 0 && k x.key(i)) ( i++; ) ) _Insert(x.child(i), k); ) ) public void Show() ( Show(root); ) private void Remove(Node x, int key) ( int pos = x.Find(key); if (pos != -1) ( if (x.leaf) ( int i = 0; for (i = 0; i < x.n && x.key(i) != key; i++) ( ) ; for (; i = T) ( for (;;) ( if (pred.leaf) ( System.out.println(pred.n); predKey = pred.key(pred.n - 1); break; ) else ( pred = pred.child(pred.n); ) ) Remove(pred, predKey); x.key(pos) = predKey; return; ) Node nextNode = x.child(pos + 1); if (nextNode.n>= T) ( int nextKey = nextNode.key(0); if (!nextNode.leaf) ( nextNode = nextNode.child(0); for (;;) ( if (nextNode.leaf) ( nextKey = nextNode.key(nextNode.n - 1); break; ) else ( nextNode = nextNode.child(nextNode.n); ) ) ) Remove(nextNode, nextKey); x.key(pos) = nextKey; return; ) int temp = pred.n + 1; pred.key(pred.n++) = x.key(pos); for (int i = 0, j = pred.n; i < nextNode.n; i++) ( pred.key(j++) = nextNode.key(i); pred.n++; ) for (int i = 0; i < nextNode.n + 1; i++) ( pred.child(temp++) = nextNode.child(i); ) x.child(pos) = pred; for (int i = pos; i < x.n; i++) ( if (i != 2 * T - 2) ( x.key(i) = x.key(i + 1); ) ) for (int i = pos + 1; i < x.n + 1; i++) ( if (i != 2 * T - 1) ( x.child(i) = x.child(i + 1); ) ) x.n--; if (x.n == 0) ( if (x == root) ( root = x.child(0); ) x = x.child(0); ) Remove(pred, key); return; ) ) else ( for (pos = 0; pos key) ( break; ) ) Node tmp = x.child(pos); if (tmp.n>= T) ( Remove(tmp, key); return; ) if (true) ( Node nb = null; int devider = -1; if (pos != x.n && x.child(pos + 1).n>= T) ( devider = x.key(pos); nb = x.child(pos + 1); x.key(pos) = nb.key(0); tmp.key(tmp.n++) = devider; tmp.child(tmp.n) = nb.child(0); for (int i = 1; i < nb.n; i++) ( nb.key(i - 1) = nb.key(i); ) for (int i = 1; i = T) ( devider = x.key(pos - 1); nb = x.child(pos - 1); x.key(pos - 1) = nb.key(nb.n - 1); Node child = nb.child(nb.n); nb.n--; for (int i = tmp.n; i> 0; i--) ( tmp.key(i) = tmp.key(i - 1); ) tmp.key(0) = devider; for (int i = tmp.n + 1; i> 0; i--) ( tmp.child(i) = tmp.child(i - 1); ) tmp.child(0) = child; tmp.n++; Remove(tmp, key); return; ) else ( Node lt = null; Node rt = null; boolean last = false; if (pos != x.n) ( devider = x.key(pos); lt = x.child(pos); rt = x.child(pos + 1); ) else ( devider = x.key(pos - 1); rt = x.child(pos); lt = x.child(pos - 1); last = true; pos--; ) for (int i = pos; i < x.n - 1; i++) ( x.key(i) = x.key(i + 1); ) for (int i = pos + 1; i < x.n; i++) ( x.child(i) = x.child(i + 1); ) x.n--; lt.key(lt.n++) = devider; for (int i = 0, j = lt.n; i < rt.n + 1; i++, j++) ( if (i < rt.n) ( lt.key(j) = rt.key(i); ) lt.child(j) = rt.child(i); ) lt.n += rt.n; if (x.n == 0) ( if (x == root) ( root = x.child(0); ) x = x.child(0); ) Remove(lt, key); return; ) ) ) ) public void Remove(int key) ( Node x = Search(root, key); if (x == null) ( return; ) Remove(root, key); ) public void Task(int a, int b) ( Stack st = new Stack(); FindKeys(a, b, root, st); while (st.isEmpty() == false) ( this.Remove(root, st.pop()); ) ) private void FindKeys(int a, int b, Node x, Stack st) ( int i = 0; for (i = 0; i < x.n && x.key(i) a) ( st.push(x.key(i)); ) ) if (!x.leaf) ( for (int j = 0; j < i + 1; j++) ( FindKeys(a, b, x.child(j), st); ) ) ) public boolean Contain(int k) ( if (this.Search(root, k) != null) ( return true; ) else ( return false; ) ) // Show the node private void Show(Node x) ( assert (x == null); for (int i = 0; i < x.n; i++) ( System.out.print(x.key(i) + " "); ) if (!x.leaf) ( for (int i = 0; i < x.n + 1; i++) ( Show(x.child(i)); ) ) ) public static void main(String() args) ( BTree b = new BTree(3); b.Insert(8); b.Insert(9); b.Insert(10); b.Insert(11); b.Insert(15); b.Insert(20); b.Insert(17); b.Show(); b.Remove(10); System.out.println(); b.Show(); ) ) 
 // Deleting a key from a B-tree in C #include #include #define MAX 3 #define MIN 2 struct BTreeNode ( int item(MAX + 1), count; struct BTreeNode *linker(MAX + 1); ); struct BTreeNode *root; // Node creation struct BTreeNode *createNode(int item, struct BTreeNode *child) ( struct BTreeNode *newNode; newNode = (struct BTreeNode *)malloc(sizeof(struct BTreeNode)); newNode->item(1) = item; newNode->count = 1; newNode->linker(0) = root; newNode->linker(1) = child; return newNode; ) // Add value to the node void addValToNode(int item, int pos, struct BTreeNode *node, struct BTreeNode *child) ( int j = node->count; while (j> pos) ( node->item(j + 1) = node->item(j); node->linker(j + 1) = node->linker(j); j--; ) node->item(j + 1) = item; node->linker(j + 1) = child; node->count++; ) // Split the node void splitNode(int item, int *pval, int pos, struct BTreeNode *node, struct BTreeNode *child, struct BTreeNode **newNode) ( int median, j; if (pos> MIN) median = MIN + 1; else median = MIN; *newNode = (struct BTreeNode *)malloc(sizeof(struct BTreeNode)); j = median + 1; while (j item(j - median) = node->item(j); (*newNode)->linker(j - median) = node->linker(j); j++; ) node->count = median; (*newNode)->count = MAX - median; if (pos item(node->count); (*newNode)->linker(0) = node->linker(node->count); node->count--; ) // Set the value in the node int setValueInNode(int item, int *pval, struct BTreeNode *node, struct BTreeNode **child) ( int pos; if (!node) ( *pval = item; *child = NULL; return 1; ) if (item item(1)) ( pos = 0; ) else ( for (pos = node->count; (item item(pos) && pos> 1); pos--) ; if (item == node->item(pos)) ( printf("Duplicates not allowed"); return 0; ) ) if (setValueInNode(item, pval, node->linker(pos), child)) ( if (node->count linker(pos); for (; dummy->linker(0) != NULL;) dummy = dummy->linker(0); myNode->item(pos) = dummy->item(1); ) // Remove the value void removeVal(struct BTreeNode *myNode, int pos) ( int i = pos + 1; while (i count) ( myNode->item(i - 1) = myNode->item(i); myNode->linker(i - 1) = myNode->linker(i); i++; ) myNode->count--; ) // Do right shift void rightShift(struct BTreeNode *myNode, int pos) ( struct BTreeNode *x = myNode->linker(pos); int j = x->count; while (j> 0) ( x->item(j + 1) = x->item(j); x->linker(j + 1) = x->linker(j); ) x->item(1) = myNode->item(pos); x->linker(1) = x->linker(0); x->count++; x = myNode->linker(pos - 1); myNode->item(pos) = x->item(x->count); myNode->linker(pos) = x->linker(x->count); x->count--; return; ) // Do left shift void leftShift(struct BTreeNode *myNode, int pos) ( int j = 1; struct BTreeNode *x = myNode->linker(pos - 1); x->count++; x->item(x->count) = myNode->item(pos); x->linker(x->count) = myNode->linker(pos)->linker(0); x = myNode->linker(pos); myNode->item(pos) = x->item(1); x->linker(0) = x->linker(1); x->count--; while (j count) ( x->item(j) = x->item(j + 1); x->linker(j) = x->linker(j + 1); j++; ) return; ) // Merge the nodes void mergeNodes(struct BTreeNode *myNode, int pos) ( int j = 1; struct BTreeNode *x1 = myNode->linker(pos), *x2 = myNode->linker(pos - 1); x2->count++; x2->item(x2->count) = myNode->item(pos); x2->linker(x2->count) = myNode->linker(0); while (j count) ( x2->count++; x2->item(x2->count) = x1->item(j); x2->linker(x2->count) = x1->linker(j); j++; ) j = pos; while (j count) ( myNode->item(j) = myNode->item(j + 1); myNode->linker(j) = myNode->linker(j + 1); j++; ) myNode->count--; free(x1); ) // Adjust the node void adjustNode(struct BTreeNode *myNode, int pos) ( if (!pos) ( if (myNode->linker(1)->count> MIN) ( leftShift(myNode, 1); ) else ( mergeNodes(myNode, 1); ) ) else ( if (myNode->count != pos) ( if (myNode->linker(pos - 1)->count> MIN) ( rightShift(myNode, pos); ) else ( if (myNode->linker(pos + 1)->count> MIN) ( leftShift(myNode, pos + 1); ) else ( mergeNodes(myNode, pos); ) ) ) else ( if (myNode->linker(pos - 1)->count> MIN) rightShift(myNode, pos); else mergeNodes(myNode, pos); ) ) ) // Delete a value from the node int delValFromNode(int item, struct BTreeNode *myNode) ( int pos, flag = 0; if (myNode) ( if (item item(1)) ( pos = 0; flag = 0; ) else ( for (pos = myNode->count; (item item(pos) && pos> 1); pos--) ; if (item == myNode->item(pos)) ( flag = 1; ) else ( flag = 0; ) ) if (flag) ( if (myNode->linker(pos - 1)) ( copySuccessor(myNode, pos); flag = delValFromNode(myNode->item(pos), myNode->linker(pos)); if (flag == 0) ( printf("Given data is not present in B-Tree"); ) ) else ( removeVal(myNode, pos); ) ) else ( flag = delValFromNode(item, myNode->linker(pos)); ) if (myNode->linker(pos)) ( if (myNode->linker(pos)->count count == 0) ( tmp = myNode; myNode = myNode->linker(0); free(tmp); ) ) root = myNode; return; ) void searching(int item, int *pos, struct BTreeNode *myNode) ( if (!myNode) ( return; ) if (item item(1)) ( *pos = 0; ) else ( for (*pos = myNode->count; (item item(*pos) && *pos> 1); (*pos)--) ; if (item == myNode->item(*pos)) ( printf("%d present in B-tree", item); return; ) ) searching(item, pos, myNode->linker(*pos)); return; ) void traversal(struct BTreeNode *myNode) ( int i; if (myNode) ( for (i = 0; i count; i++) ( traversal(myNode->linker(i)); printf("%d ", myNode->item(i + 1)); ) traversal(myNode->linker(i)); ) ) int main() ( int item, ch; insertion(8); insertion(9); insertion(10); insertion(11); insertion(15); insertion(16); insertion(17); insertion(18); insertion(20); insertion(23); traversal(root); delete (20, root); printf(""); traversal(root); )
 // Deleting a key from a B-tree in C++ #include using namespace std; class BTreeNode ( int *keys; int t; BTreeNode **C; int n; bool leaf; public: BTreeNode(int _t, bool _leaf); void traverse(); int findKey(int k); void insertNonFull(int k); void splitChild(int i, BTreeNode *y); void deletion(int k); void removeFromLeaf(int idx); void removeFromNonLeaf(int idx); int getPredecessor(int idx); int getSuccessor(int idx); void fill(int idx); void borrowFromPrev(int idx); void borrowFromNext(int idx); void merge(int idx); friend class BTree; ); class BTree ( BTreeNode *root; int t; public: BTree(int _t) ( root = NULL; t = _t; ) void traverse() ( if (root != NULL) root->traverse(); ) void insertion(int k); void deletion(int k); ); // B tree node BTreeNode::BTreeNode(int t1, bool leaf1) ( t = t1; leaf = leaf1; keys = new int(2 * t - 1); C = new BTreeNode *(2 * t); n = 0; ) // Find the key int BTreeNode::findKey(int k) ( int idx = 0; while (idx < n && keys(idx) < k) ++idx; return idx; ) // Deletion operation void BTreeNode::deletion(int k) ( int idx = findKey(k); if (idx < n && keys(idx) == k) ( if (leaf) removeFromLeaf(idx); else removeFromNonLeaf(idx); ) else ( if (leaf) ( cout << "The key " << k  deletion(k); else C(idx)->deletion(k); ) return; ) // Remove from the leaf void BTreeNode::removeFromLeaf(int idx) ( for (int i = idx + 1; i n>= t) ( int pred = getPredecessor(idx); keys(idx) = pred; C(idx)->deletion(pred); ) else if (C(idx + 1)->n>= t) ( int succ = getSuccessor(idx); keys(idx) = succ; C(idx + 1)->deletion(succ); ) else ( merge(idx); C(idx)->deletion(k); ) return; ) int BTreeNode::getPredecessor(int idx) ( BTreeNode *cur = C(idx); while (!cur->leaf) cur = cur->C(cur->n); return cur->keys(cur->n - 1); ) int BTreeNode::getSuccessor(int idx) ( BTreeNode *cur = C(idx + 1); while (!cur->leaf) cur = cur->C(0); return cur->keys(0); ) void BTreeNode::fill(int idx) ( if (idx != 0 && C(idx - 1)->n>= t) borrowFromPrev(idx); else if (idx != n && C(idx + 1)->n>= t) borrowFromNext(idx); else ( if (idx != n) merge(idx); else merge(idx - 1); ) return; ) // Borrow from previous void BTreeNode::borrowFromPrev(int idx) ( BTreeNode *child = C(idx); BTreeNode *sibling = C(idx - 1); for (int i = child->n - 1; i>= 0; --i) child->keys(i + 1) = child->keys(i); if (!child->leaf) ( for (int i = child->n; i>= 0; --i) child->C(i + 1) = child->C(i); ) child->keys(0) = keys(idx - 1); if (!child->leaf) child->C(0) = sibling->C(sibling->n); keys(idx - 1) = sibling->keys(sibling->n - 1); child->n += 1; sibling->n -= 1; return; ) // Borrow from the next void BTreeNode::borrowFromNext(int idx) ( BTreeNode *child = C(idx); BTreeNode *sibling = C(idx + 1); child->keys((child->n)) = keys(idx); if (!(child->leaf)) child->C((child->n) + 1) = sibling->C(0); keys(idx) = sibling->keys(0); for (int i = 1; i n; ++i) sibling->keys(i - 1) = sibling->keys(i); if (!sibling->leaf) ( for (int i = 1; i n; ++i) sibling->C(i - 1) = sibling->C(i); ) child->n += 1; sibling->n -= 1; return; ) // Merge void BTreeNode::merge(int idx) ( BTreeNode *child = C(idx); BTreeNode *sibling = C(idx + 1); child->keys(t - 1) = keys(idx); for (int i = 0; i n; ++i) child->keys(i + t) = sibling->keys(i); if (!child->leaf) ( for (int i = 0; i n; ++i) child->C(i + t) = sibling->C(i); ) for (int i = idx + 1; i < n; ++i) keys(i - 1) = keys(i); for (int i = idx + 2; i n += sibling->n + 1; n--; delete (sibling); return; ) // Insertion operation void BTree::insertion(int k) ( if (root == NULL) ( root = new BTreeNode(t, true); root->keys(0) = k; root->n = 1; ) else ( if (root->n == 2 * t - 1) ( BTreeNode *s = new BTreeNode(t, false); s->C(0) = root; s->splitChild(0, root); int i = 0; if (s->keys(0) C(i)->insertNonFull(k); root = s; ) else root->insertNonFull(k); ) ) // Insertion non full void BTreeNode::insertNonFull(int k) ( int i = n - 1; if (leaf == true) ( while (i>= 0 && keys(i)> k) ( keys(i + 1) = keys(i); i--; ) keys(i + 1) = k; n = n + 1; ) else ( while (i>= 0 && keys(i)> k) i--; if (C(i + 1)->n == 2 * t - 1) ( splitChild(i + 1, C(i + 1)); if (keys(i + 1) insertNonFull(k); ) ) // Split child void BTreeNode::splitChild(int i, BTreeNode *y) ( BTreeNode *z = new BTreeNode(y->t, y->leaf); z->n = t - 1; for (int j = 0; j keys(j) = y->keys(j + t); if (y->leaf == false) ( for (int j = 0; j C(j) = y->C(j + t); ) y->n = t - 1; for (int j = n; j>= i + 1; j--) C(j + 1) = C(j); C(i + 1) = z; for (int j = n - 1; j>= i; j--) keys(j + 1) = keys(j); keys(i) = y->keys(t - 1); n = n + 1; ) // Traverse void BTreeNode::traverse() ( int i; for (i = 0; i traverse(); cout << " " 
 n == 0) ( BTreeNode *tmp = root; if (root->leaf) root = NULL; else root = root->C(0); delete tmp; ) return; ) int main() ( BTree t(3); t.insertion(8); t.insertion(9); t.insertion(10); t.insertion(11); t.insertion(15); t.insertion(16); t.insertion(17); t.insertion(18); t.insertion(20); t.insertion(23); cout << "The B-tree is: "; t.traverse(); t.deletion(20); cout << "The B-tree is: "; t.traverse(); )

Dzēšanas sarežģītība

Labākais gadījuma sarežģītība: Θ(log n)

Vidējā telpa sarežģītība: Θ(n)

Sliktākais gadījums - kosmiskā sarežģītība: Θ(n)

Interesanti raksti...