Bartuccio Antoine
48e43c10eb
All checks were successful
continuous-integration/drone/push Build is passing
497 lines
12 KiB
Go
497 lines
12 KiB
Go
package btree
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import (
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"fmt"
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)
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// Tree is the tree itself
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type Tree struct {
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root *Node // Pointer to the Node root
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t int // Minimum degree
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}
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// Node is a Node of a Btree
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type Node struct {
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numberOfKeys int // The number of keys really stored
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degree int // The value of degree dependes upon disk blok size
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isLeaf bool
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keys []int
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children []*Node
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}
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// Constructors
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// NewBtree creates a new btree
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func NewBtree(t int) *Tree {
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return &Tree{
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root: nil,
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t: t,
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}
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}
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func newNode(degree int, isLeaf bool) *Node {
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return &Node{
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numberOfKeys: 0,
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degree: degree,
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isLeaf: isLeaf,
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keys: make([]int, 2*degree-1),
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children: make([]*Node, 2*degree),
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}
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}
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// Tree methods
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// Traverse the tree
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func (t *Tree) Traverse() {
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if t.root != nil {
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t.root.traverse()
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}
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}
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// Search k in the tree
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func (t *Tree) Search(k int) *Node {
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if t.root == nil {
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return nil
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}
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return t.root.search(k)
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}
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// Remove k in the tree
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func (t *Tree) Remove(k int) error {
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if t.root == nil {
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return fmt.Errorf("The tree is empty")
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}
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err := t.root.remove(k)
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// If the root node has 0 keys, makes its first child as the new root
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// If it has no child, set root as nil
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if t.root.numberOfKeys == 0 {
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if t.root.isLeaf {
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t.root = nil
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return err
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}
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t.root = t.root.children[0]
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}
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return err
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}
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// Insert k in the tree
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func (t *Tree) Insert(k int) {
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// If the tree is empty
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if t.root == nil {
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t.root = newNode(t.t, true)
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t.root.keys[0] = k
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t.root.numberOfKeys = 1
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return
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}
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// If the tree is not empty
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if !t.root.isFull() {
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// If root is not full, insert in non full root
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t.root.insertNonFull(k)
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return
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}
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// If the root is full, then the tree grows in height
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s := newNode(t.t, false)
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// Make the old root as a child of the new root
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s.children[0] = t.root
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// Split the old root and move 1 key to the new root
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s.splitChild(0, t.root)
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// The new root has two children now.
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// We decide which of the two children is going to have the new key
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i := 0
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if s.keys[0] < k {
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i++
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}
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s.children[i].insertNonFull(k)
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// Change root
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t.root = s
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}
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// Node methods
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// traverse all nodes in a subtree rooted with this node
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func (n *Node) traverse() {
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// There are n entries and n+1 children, treverse trough n keys and n first children
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for i := 0; i < n.numberOfKeys; i++ {
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// If this is not a leaf, then traverse the subtree before printing the key
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if !n.isLeaf {
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n.children[i].traverse()
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}
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fmt.Printf(" %d", n.keys[i])
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}
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// Print the subtree rooted with the last child
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if !n.isLeaf {
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n.children[n.numberOfKeys].traverse()
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}
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}
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// search k in the subtree rooted with this node
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func (n *Node) search(k int) *Node {
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// Find the first entry greater than or equal to k
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i := 0
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for i < n.numberOfKeys && k > n.keys[i] {
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i++
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}
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// If theh found key is equal to k, return this node
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if n.keys[i] == k {
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return n
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}
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// If the key is not found here and this is a leaf node
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if n.isLeaf {
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return nil
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}
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// Go to the approipriate child
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return n.children[i].search(k)
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}
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func (n *Node) isFull() bool {
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return n.numberOfKeys == 2*n.degree-1
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}
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func (n *Node) insertNonFull(k int) {
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// Initialize the index as the index of the rightmost element
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i := n.numberOfKeys - 1
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// If this is a leaf node
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if n.isLeaf {
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// Finds the location of the new key to be inserted
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// Moves all greater keys to one place ahead
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for i >= 0 && n.keys[i] > k {
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n.keys[i+1] = n.keys[i]
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i--
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}
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// Insert the new key at the found location
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n.keys[i+1] = k
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n.numberOfKeys++
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return
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}
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// If this is not a leaf
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// Finds the child wich is going to have the new key
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for i >= 0 && n.keys[i] > k {
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i--
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}
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// Check if the found chird is full
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if n.children[i+1].isFull() {
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// If the child is full, then split it
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n.splitChild(i+1, n.children[i+1])
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// After the split, the middle key of children[i] goes up and
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// children[i] is splitted into two
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// See which of those two is going to have the new key
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if n.keys[i+1] < k {
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i++
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}
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}
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n.children[i+1].insertNonFull(k)
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}
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func (n *Node) splitChild(i int, y *Node) {
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// Create a new node that will store (t-1) keys of y
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z := newNode(y.degree, y.isLeaf)
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z.numberOfKeys = n.degree - 1
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// Copy the last (t-1) keys of y to z
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for j := 0; j < n.degree-1; j++ {
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z.keys[j] = y.keys[j+n.degree]
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if !y.isLeaf {
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z.children[j] = y.children[j+n.degree]
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}
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}
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// Copy the last t children of y to z
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if !y.isLeaf {
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z.children[n.degree-1] = y.children[2*n.degree-1]
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}
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// Reduce the number of keys in y
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y.numberOfKeys = n.degree - 1
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// Since this node is going to have a new child, create space for it
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for j := n.numberOfKeys; j >= i+1; j-- {
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n.children[j+1] = n.children[j]
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}
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// Link the new child to this node
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n.children[i+1] = z
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// A key of y will move to this node
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// Find the location of the new key and move all greater keys ahead
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for j := n.numberOfKeys - 1; j >= i; j-- {
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n.keys[j+1] = n.keys[j]
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}
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// Copy the middle key of y to this node
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n.keys[i] = y.keys[n.degree-1]
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// Increment the count of keys in this node
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n.numberOfKeys++
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}
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// findKey returns the index of the first key that is greater than or equal to k
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func (n *Node) findKey(k int) int {
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index := 0
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for index < n.numberOfKeys && n.keys[index] < k {
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index++
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}
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return index
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}
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// remove the key k from the sub-tree rooted with this node
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func (n *Node) remove(k int) error {
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index := n.findKey(k)
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// The key to be removed is in this node
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if index < n.numberOfKeys && n.keys[index] == k {
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if n.isLeaf {
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return n.removeFromLeaf(index)
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}
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return n.removeFromNonLeaf(index)
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}
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// If this is a leaf, the key is not in the tree
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if n.isLeaf {
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return fmt.Errorf("The key %d does not exist in the tree", k)
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}
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isInLastChild := false
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if index == n.numberOfKeys {
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isInLastChild = true
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}
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// If the child where is the key has less than t keys, wi fill it
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if n.children[index].numberOfKeys < n.degree {
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n.fill(index)
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}
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// If the last child has been merged, it must be merged with the previous
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// child and so we recurse on the (index-1)th child.
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if isInLastChild && index > n.numberOfKeys {
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return n.children[index-1].remove(k)
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}
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// We recurse on the (index)th child which now has at least t keys
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return n.children[index].remove(k)
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}
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// removeFromLeaf the index-th key from this node which is a leaf node
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func (n *Node) removeFromLeaf(index int) error {
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// Move all the keys after the index-th position one place backward
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for i := index + 1; i < n.numberOfKeys; i++ {
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n.keys[i-1] = n.keys[i]
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}
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n.numberOfKeys--
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return nil
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}
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// removeFromNonLeaf the index-th key from this node which is not a leaf node
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func (n *Node) removeFromNonLeaf(index int) error {
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k := n.keys[index]
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// If the child that precedes k has at least t keys,
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// find the predecessor of k in the subtree and replace k with it
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// Recursively delete the predecessor in the child
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if n.children[index].numberOfKeys >= n.degree {
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pred := n.getPred(index)
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n.keys[index] = pred
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return n.children[index].remove(pred)
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}
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// If the child has less than t keys, examine children[index+1]
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// If it has at least t keys, find the successor of k in this subtree
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// Replace k by its successor and recursively delete the successor in the subtree
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if n.children[index+1].numberOfKeys >= n.degree {
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succ := n.getSucc(index)
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n.keys[index] = succ
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return n.children[index+1].remove(succ)
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}
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// Merge k and all of children[index+1] int children[index]
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// Free children[index+1] and recursively delete k from children[index]
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n.merge(index)
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return n.children[index].remove(k)
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}
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// getPred returns the predecessor of keys[index]
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func (n *Node) getPred(index int) int {
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// Keep moving to the rightmost node until we reach a leaf
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current := n.children[index]
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for !current.isLeaf {
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current = current.children[current.numberOfKeys]
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}
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// Return the last key of the leaf
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return current.keys[current.numberOfKeys-1]
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}
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// getSucc returns the successor of keys[index]
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func (n *Node) getSucc(index int) int {
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// Keep moving to the leftmost node starting from children[index+1] until we reach a leaf
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current := n.children[index+1]
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for !current.isLeaf {
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current = current.children[0]
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}
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// Return the first key of the leaf
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return current.keys[0]
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}
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// fill child children[index] which has less than t-1 keys
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func (n *Node) fill(index int) {
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// If the previous child has more than t-1 keys, borrow a key from that child
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if index != 0 && n.children[index-1].numberOfKeys >= n.degree {
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n.borrowFromPrev(index)
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return
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}
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// If the next child has more than t-1 keys, borrow a key from that child
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if index != n.numberOfKeys && n.children[index+1].numberOfKeys >= n.degree {
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n.borrowFromNext(index)
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return
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}
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// Merge children[index] with its sibling
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if index != n.numberOfKeys {
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n.merge(index)
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return
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}
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// If this is the last child, merge with the previous sibling
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n.merge(index - 1)
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}
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// borrowFromPrev takes a key from children[index+1] and insert it in children[index]
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func (n *Node) borrowFromPrev(index int) {
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child := n.children[index]
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sibling := n.children[index-1]
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// Moves all keys in children[index] one step ahead
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for i := child.numberOfKeys - 1; i >= 0; i-- {
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child.keys[i+1] = child.keys[i]
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}
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// If the child is not a leaf, move all its child pointers one step ahead
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if !child.isLeaf {
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for i := child.numberOfKeys; i >= 0; i-- {
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child.children[i+1] = child.children[i]
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}
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}
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// Sets child's first key equal to keys[index-1] from the current node
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child.keys[0] = n.keys[index-1]
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// Moves sibling's last child as children[index]'s first child
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if !child.isLeaf {
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child.children[0] = sibling.children[sibling.numberOfKeys]
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}
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// Moves the key from the sibling to the parent
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n.keys[index-1] = sibling.keys[sibling.numberOfKeys-1]
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child.numberOfKeys++
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sibling.numberOfKeys--
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}
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// borrowFromNext takes a key from children[index+1] and insert it in children[index]
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func (n *Node) borrowFromNext(index int) {
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child := n.children[index]
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sibling := n.children[index+1]
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// keys[index] is inserted as the last key in children[index]
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child.keys[child.numberOfKeys] = n.keys[index]
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// Sibling's first child is inserted as the last child into children[index]
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if !child.isLeaf {
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child.children[child.numberOfKeys+1] = sibling.children[0]
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}
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// The first key from sibling is inserted into keys[index]
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n.keys[index] = sibling.keys[0]
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// Moving all keys in sibling one step behind
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for i := 1; i < sibling.numberOfKeys; i++ {
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sibling.keys[i-1] = sibling.keys[i]
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if !sibling.isLeaf {
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sibling.children[i-1] = sibling.children[i]
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}
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}
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// Moving the child pointers one step behind
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if !sibling.isLeaf {
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sibling.children[n.numberOfKeys-1] = sibling.children[n.numberOfKeys]
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}
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child.numberOfKeys++
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sibling.numberOfKeys--
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}
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// merge children[index] with children[index+1]
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func (n *Node) merge(index int) {
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child := n.children[index]
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sibling := n.children[index+1]
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// Pulls a key from the current node ande insert it into the (t-1)th position
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child.keys[n.degree-1] = n.keys[index]
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// Copies the keys from children[index+1] to children[index] at the end
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for i := 0; i < sibling.numberOfKeys; i++ {
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child.keys[i+n.degree] = sibling.keys[i]
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if !child.isLeaf {
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child.children[i+n.degree] = sibling.children[i]
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}
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}
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// Copies the child pointers from C[index+1] to children[index]
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if !child.isLeaf {
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child.children[sibling.numberOfKeys+n.degree] = sibling.children[sibling.numberOfKeys]
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}
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// Moves all keys after index in the current node one step before
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// to fill the gap created by moving keys[index] to children[index]
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// Moves the child pointer after (index+1) in the current node one step before
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// This action marks sibling for deletion by the GC
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for i := index + 1; i < n.numberOfKeys; i++ {
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n.keys[i-1] = n.keys[i]
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n.children[i] = n.children[i+1]
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}
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// Updates the key count of child and the current node
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child.numberOfKeys += sibling.numberOfKeys + 1
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n.numberOfKeys--
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}
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