private void fixAfterDeletion(int parentIndex) {
if (isRoot() || parent.isRoot()) return; // No fixing needed
if (parent.nrElements < MIN) { // If parent lost it's n/2 element repair it
BTreeNode temp = parent;
temp.prepareForDeletion(parentIndex);
if (temp.parent == null) return; // Root changed
if (!temp.parent.isRoot() && temp.parent.nrElements < MIN) { // If need be recurse
BTreeNode x = temp.parent.parent;
int i = 0;
// Find parent's parentIndex
for (; i < entries.length; i++) if (x.entries[i].child == temp.parent) break;
temp.parent.fixAfterDeletion(i);
}
}
}
/*
* This method is called only when stealLeft, stealRight, and mergeLeft could not be called,
* the BTreeNode has the minimum number of elements, has a rightSibling, and the
* rightSibling has more than the minimum number of elements. If after completion
* parent has fewer than the minimum number of elements than the parents entries[0]
* slot is left empty in anticipation of a recursive call to stealLeft, stealRight,
* mergeLeft, or mergeRight to fix the parent. All of the before-mentioned methods
* expect the parent to be in such a condition.
*/
private void mergeRight(int parentIndex) {
BTreeNode p = parent;
BTreeNode rs = p.entries[parentIndex + 1].child;
if (isLeaf()) { // Don't worry about children
entries[nrElements] = new Entry();
entries[nrElements].element = p.entries[parentIndex].element;
nrElements++;
for (int i = 0, nr = nrElements; i < rs.nrElements; i++, nr++) {
entries[nr] = rs.entries[i];
nrElements++;
}
p.entries[parentIndex].element = p.entries[parentIndex + 1].element;
if (p.nrElements == MIN && p != BTreeSet.this.root) {
for (int x = parentIndex + 1, y = parentIndex; y >= 0; x--, y--)
p.entries[x] = p.entries[y];
p.entries[0] = new Entry();
p.entries[0].child =
rs; // So it doesn't think it's a leaf, this child will be deleted in the next
// recursive call
} else {
for (int x = parentIndex + 1, y = parentIndex + 2; y <= p.nrElements; x++, y++)
p.entries[x] = p.entries[y];
p.entries[p.nrElements] = null;
}
p.nrElements--;
if (p.isRoot() && p.nrElements == 0) { // It's the root and it's empty
BTreeSet.this.root = this;
parent = null;
}
} else { // It's not a leaf
entries[nrElements].element = p.entries[parentIndex].element;
nrElements++;
for (int x = nrElements + 1, y = 0; y <= rs.nrElements; x++, y++) {
entries[x] = rs.entries[y];
rs.entries[y].child.parent = this;
nrElements++;
}
nrElements--;
p.entries[++parentIndex].child = this;
if (p.nrElements == MIN && p != BTreeSet.this.root) {
for (int x = parentIndex - 1, y = parentIndex - 2; y >= 0; x--, y--)
p.entries[x] = p.entries[y];
p.entries[0] = new Entry();
} else {
for (int x = parentIndex - 1, y = parentIndex; y <= p.nrElements; x++, y++)
p.entries[x] = p.entries[y];
p.entries[p.nrElements] = null;
}
p.nrElements--;
if (p.isRoot() && p.nrElements == 0) { // It's the root and it's empty
BTreeSet.this.root = this;
parent = null;
}
}
}
/*
* This method is called only when stealLeft and stealRight could not be called,
* the BTreeNode has the minimum number of elements, has a leftSibling, and the
* leftSibling has more than the minimum number of elements. If after completion
* parent has fewer than the minimum number of elements than the parents entries[0]
* slot is left empty in anticipation of a recursive call to stealLeft, stealRight,
* mergeLeft, or mergeRight to fix the parent. All of the before-mentioned methods
* expect the parent to be in such a condition.
*/
private void mergeLeft(int parentIndex) {
BTreeNode p = parent;
BTreeNode ls = p.entries[parentIndex - 1].child;
if (isLeaf()) { // Don't worry about children
int add = childToInsertAt(p.entries[parentIndex - 1].element, true);
insertNewElement(
p.entries[parentIndex - 1].element, add); // Could have been a successor switch
p.entries[parentIndex - 1].element = null;
for (int i = nrElements - 1, nr = ls.nrElements; i >= 0; i--) entries[i + nr] = entries[i];
for (int i = ls.nrElements - 1; i >= 0; i--) {
entries[i] = ls.entries[i];
nrElements++;
}
if (p.nrElements == MIN && p != BTreeSet.this.root) {
for (int x = parentIndex - 1, y = parentIndex - 2; y >= 0; x--, y--)
p.entries[x] = p.entries[y];
p.entries[0] = new Entry();
p.entries[0].child =
ls; // So p doesn't think it's a leaf this will be deleted in the next recursive call
} else {
for (int x = parentIndex - 1, y = parentIndex; y <= p.nrElements; x++, y++)
p.entries[x] = p.entries[y];
p.entries[p.nrElements] = null;
}
p.nrElements--;
if (p.isRoot() && p.nrElements == 0) { // It's the root and it's empty
BTreeSet.this.root = this;
parent = null;
}
} else { // I'm not a leaf but fixing the tree structure
entries[0].element = p.entries[parentIndex - 1].element;
entries[0].child = ls.entries[ls.nrElements].child;
nrElements++;
for (int x = nrElements, nr = ls.nrElements; x >= 0; x--) entries[x + nr] = entries[x];
for (int x = ls.nrElements - 1; x >= 0; x--) {
entries[x] = ls.entries[x];
entries[x].child.parent = this;
nrElements++;
}
if (p.nrElements == MIN && p != BTreeSet.this.root) { // Push everything to the right
for (int x = parentIndex - 1, y = parentIndex - 2; y >= 0; x++, y++) {
System.out.println(x + " " + y);
p.entries[x] = p.entries[y];
}
p.entries[0] = new Entry();
} else { // Either p.nrElements > MIN or p == BTreeSet.this.root so push everything to the
// left
for (int x = parentIndex - 1, y = parentIndex; y <= p.nrElements; x++, y++)
p.entries[x] = p.entries[y];
p.entries[p.nrElements] = null;
}
p.nrElements--;
if (p.isRoot() && p.nrElements == 0) { // p == BTreeSet.this.root and it's empty
BTreeSet.this.root = this;
parent = null;
}
}
}