2023-08-11 13:12:13 +00:00
|
|
|
// Copyright (c) HashiCorp, Inc.
|
|
|
|
// SPDX-License-Identifier: BUSL-1.1
|
|
|
|
|
2023-06-06 21:09:48 +00:00
|
|
|
package radix
|
|
|
|
|
|
|
|
import (
|
|
|
|
"sort"
|
|
|
|
"strings"
|
|
|
|
)
|
|
|
|
|
|
|
|
// WalkFn is used when walking the tree. Takes a
|
|
|
|
// key and value, returning if iteration should
|
|
|
|
// be terminated.
|
|
|
|
type WalkFn[T any] func(s string, v T) bool
|
|
|
|
|
|
|
|
// leafNode is used to represent a value
|
|
|
|
type leafNode[T any] struct {
|
|
|
|
key string
|
|
|
|
val T
|
|
|
|
}
|
|
|
|
|
|
|
|
// edge is used to represent an edge node
|
|
|
|
type edge[T any] struct {
|
|
|
|
label byte
|
|
|
|
node *node[T]
|
|
|
|
}
|
|
|
|
|
|
|
|
type node[T any] struct {
|
|
|
|
// leaf is used to store possible leaf
|
|
|
|
leaf *leafNode[T]
|
|
|
|
|
|
|
|
// prefix is the common prefix we ignore
|
|
|
|
prefix string
|
|
|
|
|
|
|
|
// Edges should be stored in-order for iteration.
|
|
|
|
// We avoid a fully materialized slice to save memory,
|
|
|
|
// since in most cases we expect to be sparse
|
|
|
|
edges edges[T]
|
|
|
|
}
|
|
|
|
|
|
|
|
func (n *node[T]) isLeaf() bool {
|
|
|
|
return n.leaf != nil
|
|
|
|
}
|
|
|
|
|
|
|
|
func (n *node[T]) addEdge(e edge[T]) {
|
|
|
|
num := len(n.edges)
|
|
|
|
idx := sort.Search(num, func(i int) bool {
|
|
|
|
return n.edges[i].label >= e.label
|
|
|
|
})
|
|
|
|
|
|
|
|
n.edges = append(n.edges, edge[T]{})
|
|
|
|
copy(n.edges[idx+1:], n.edges[idx:])
|
|
|
|
n.edges[idx] = e
|
|
|
|
}
|
|
|
|
|
|
|
|
func (n *node[T]) updateEdge(label byte, node *node[T]) {
|
|
|
|
num := len(n.edges)
|
|
|
|
idx := sort.Search(num, func(i int) bool {
|
|
|
|
return n.edges[i].label >= label
|
|
|
|
})
|
|
|
|
if idx < num && n.edges[idx].label == label {
|
|
|
|
n.edges[idx].node = node
|
|
|
|
return
|
|
|
|
}
|
|
|
|
panic("replacing missing edge")
|
|
|
|
}
|
|
|
|
|
|
|
|
func (n *node[T]) getEdge(label byte) *node[T] {
|
|
|
|
num := len(n.edges)
|
|
|
|
idx := sort.Search(num, func(i int) bool {
|
|
|
|
return n.edges[i].label >= label
|
|
|
|
})
|
|
|
|
if idx < num && n.edges[idx].label == label {
|
|
|
|
return n.edges[idx].node
|
|
|
|
}
|
|
|
|
return nil
|
|
|
|
}
|
|
|
|
|
|
|
|
func (n *node[T]) delEdge(label byte) {
|
|
|
|
num := len(n.edges)
|
|
|
|
idx := sort.Search(num, func(i int) bool {
|
|
|
|
return n.edges[i].label >= label
|
|
|
|
})
|
|
|
|
if idx < num && n.edges[idx].label == label {
|
|
|
|
copy(n.edges[idx:], n.edges[idx+1:])
|
|
|
|
n.edges[len(n.edges)-1] = edge[T]{}
|
|
|
|
n.edges = n.edges[:len(n.edges)-1]
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
type edges[T any] []edge[T]
|
|
|
|
|
|
|
|
func (e edges[T]) Len() int {
|
|
|
|
return len(e)
|
|
|
|
}
|
|
|
|
|
|
|
|
func (e edges[T]) Less(i, j int) bool {
|
|
|
|
return e[i].label < e[j].label
|
|
|
|
}
|
|
|
|
|
|
|
|
func (e edges[T]) Swap(i, j int) {
|
|
|
|
e[i], e[j] = e[j], e[i]
|
|
|
|
}
|
|
|
|
|
|
|
|
func (e edges[T]) Sort() {
|
|
|
|
sort.Sort(e)
|
|
|
|
}
|
|
|
|
|
|
|
|
// Tree implements a radix tree. This can be treated as a
|
|
|
|
// Dictionary abstract data type. The main advantage over
|
|
|
|
// a standard hash map is prefix-based lookups and
|
|
|
|
// ordered iteration,
|
|
|
|
type Tree[T any] struct {
|
|
|
|
root *node[T]
|
|
|
|
size int
|
|
|
|
}
|
|
|
|
|
|
|
|
// New returns an empty Tree
|
|
|
|
func New[T any]() *Tree[T] {
|
|
|
|
return NewFromMap[T](nil)
|
|
|
|
}
|
|
|
|
|
|
|
|
// NewFromMap returns a new tree containing the keys
|
|
|
|
// from an existing map
|
|
|
|
func NewFromMap[T any](m map[string]T) *Tree[T] {
|
|
|
|
t := &Tree[T]{root: &node[T]{}}
|
|
|
|
for k, v := range m {
|
|
|
|
t.Insert(k, v)
|
|
|
|
}
|
|
|
|
return t
|
|
|
|
}
|
|
|
|
|
|
|
|
// Len is used to return the number of elements in the tree
|
|
|
|
func (t *Tree[T]) Len() int {
|
|
|
|
return t.size
|
|
|
|
}
|
|
|
|
|
|
|
|
// longestPrefix finds the length of the shared prefix
|
|
|
|
// of two strings
|
|
|
|
func longestPrefix(k1, k2 string) int {
|
|
|
|
max := len(k1)
|
|
|
|
if l := len(k2); l < max {
|
|
|
|
max = l
|
|
|
|
}
|
|
|
|
var i int
|
|
|
|
for i = 0; i < max; i++ {
|
|
|
|
if k1[i] != k2[i] {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
}
|
|
|
|
return i
|
|
|
|
}
|
|
|
|
|
|
|
|
// Insert is used to add a newentry or update
|
|
|
|
// an existing entry. Returns true if an existing record is updated.
|
|
|
|
func (t *Tree[T]) Insert(s string, v T) (T, bool) {
|
|
|
|
var zeroVal T
|
|
|
|
var parent *node[T]
|
|
|
|
n := t.root
|
|
|
|
search := s
|
|
|
|
for {
|
|
|
|
// Handle key exhaution
|
|
|
|
if len(search) == 0 {
|
|
|
|
if n.isLeaf() {
|
|
|
|
old := n.leaf.val
|
|
|
|
n.leaf.val = v
|
|
|
|
return old, true
|
|
|
|
}
|
|
|
|
|
|
|
|
n.leaf = &leafNode[T]{
|
|
|
|
key: s,
|
|
|
|
val: v,
|
|
|
|
}
|
|
|
|
t.size++
|
|
|
|
return zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for the edge
|
|
|
|
parent = n
|
|
|
|
n = n.getEdge(search[0])
|
|
|
|
|
|
|
|
// No edge, create one
|
|
|
|
if n == nil {
|
|
|
|
e := edge[T]{
|
|
|
|
label: search[0],
|
|
|
|
node: &node[T]{
|
|
|
|
leaf: &leafNode[T]{
|
|
|
|
key: s,
|
|
|
|
val: v,
|
|
|
|
},
|
|
|
|
prefix: search,
|
|
|
|
},
|
|
|
|
}
|
|
|
|
parent.addEdge(e)
|
|
|
|
t.size++
|
|
|
|
return zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// Determine longest prefix of the search key on match
|
|
|
|
commonPrefix := longestPrefix(search, n.prefix)
|
|
|
|
if commonPrefix == len(n.prefix) {
|
|
|
|
search = search[commonPrefix:]
|
|
|
|
continue
|
|
|
|
}
|
|
|
|
|
|
|
|
// Split the node
|
|
|
|
t.size++
|
|
|
|
child := &node[T]{
|
|
|
|
prefix: search[:commonPrefix],
|
|
|
|
}
|
|
|
|
parent.updateEdge(search[0], child)
|
|
|
|
|
|
|
|
// Restore the existing node
|
|
|
|
child.addEdge(edge[T]{
|
|
|
|
label: n.prefix[commonPrefix],
|
|
|
|
node: n,
|
|
|
|
})
|
|
|
|
n.prefix = n.prefix[commonPrefix:]
|
|
|
|
|
|
|
|
// Create a new leaf node
|
|
|
|
leaf := &leafNode[T]{
|
|
|
|
key: s,
|
|
|
|
val: v,
|
|
|
|
}
|
|
|
|
|
|
|
|
// If the new key is a subset, add to this node
|
|
|
|
search = search[commonPrefix:]
|
|
|
|
if len(search) == 0 {
|
|
|
|
child.leaf = leaf
|
|
|
|
return zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// Create a new edge for the node
|
|
|
|
child.addEdge(edge[T]{
|
|
|
|
label: search[0],
|
|
|
|
node: &node[T]{
|
|
|
|
leaf: leaf,
|
|
|
|
prefix: search,
|
|
|
|
},
|
|
|
|
})
|
|
|
|
return zeroVal, false
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
// Delete is used to delete a key, returning the previous
|
|
|
|
// value and if it was deleted
|
|
|
|
func (t *Tree[T]) Delete(s string) (T, bool) {
|
|
|
|
var zeroVal T
|
|
|
|
|
|
|
|
var parent *node[T]
|
|
|
|
var label byte
|
|
|
|
n := t.root
|
|
|
|
search := s
|
|
|
|
for {
|
|
|
|
// Check for key exhaution
|
|
|
|
if len(search) == 0 {
|
|
|
|
if !n.isLeaf() {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
goto DELETE
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for an edge
|
|
|
|
parent = n
|
|
|
|
label = search[0]
|
|
|
|
n = n.getEdge(label)
|
|
|
|
if n == nil {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
|
|
|
|
// Consume the search prefix
|
|
|
|
if strings.HasPrefix(search, n.prefix) {
|
|
|
|
search = search[len(n.prefix):]
|
|
|
|
} else {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
}
|
|
|
|
return zeroVal, false
|
|
|
|
|
|
|
|
DELETE:
|
|
|
|
// Delete the leaf
|
|
|
|
leaf := n.leaf
|
|
|
|
n.leaf = nil
|
|
|
|
t.size--
|
|
|
|
|
|
|
|
// Check if we should delete this node from the parent
|
|
|
|
if parent != nil && len(n.edges) == 0 {
|
|
|
|
parent.delEdge(label)
|
|
|
|
}
|
|
|
|
|
|
|
|
// Check if we should merge this node
|
|
|
|
if n != t.root && len(n.edges) == 1 {
|
|
|
|
n.mergeChild()
|
|
|
|
}
|
|
|
|
|
|
|
|
// Check if we should merge the parent's other child
|
|
|
|
if parent != nil && parent != t.root && len(parent.edges) == 1 && !parent.isLeaf() {
|
|
|
|
parent.mergeChild()
|
|
|
|
}
|
|
|
|
|
|
|
|
return leaf.val, true
|
|
|
|
}
|
|
|
|
|
|
|
|
// DeletePrefix is used to delete the subtree under a prefix
|
|
|
|
// Returns how many nodes were deleted
|
|
|
|
// Use this to delete large subtrees efficiently
|
|
|
|
func (t *Tree[T]) DeletePrefix(s string) int {
|
|
|
|
return t.deletePrefix(nil, t.root, s)
|
|
|
|
}
|
|
|
|
|
|
|
|
// delete does a recursive deletion
|
|
|
|
func (t *Tree[T]) deletePrefix(parent, n *node[T], prefix string) int {
|
|
|
|
// Check for key exhaustion
|
|
|
|
if len(prefix) == 0 {
|
|
|
|
// Remove the leaf node
|
|
|
|
subTreeSize := 0
|
|
|
|
//recursively walk from all edges of the node to be deleted
|
|
|
|
recursiveWalk(n, func(s string, v T) bool {
|
|
|
|
subTreeSize++
|
|
|
|
return false
|
|
|
|
})
|
|
|
|
if n.isLeaf() {
|
|
|
|
n.leaf = nil
|
|
|
|
}
|
|
|
|
n.edges = nil // deletes the entire subtree
|
|
|
|
|
|
|
|
// Check if we should merge the parent's other child
|
|
|
|
if parent != nil && parent != t.root && len(parent.edges) == 1 && !parent.isLeaf() {
|
|
|
|
parent.mergeChild()
|
|
|
|
}
|
|
|
|
t.size -= subTreeSize
|
|
|
|
return subTreeSize
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for an edge
|
|
|
|
label := prefix[0]
|
|
|
|
child := n.getEdge(label)
|
|
|
|
if child == nil || (!strings.HasPrefix(child.prefix, prefix) && !strings.HasPrefix(prefix, child.prefix)) {
|
|
|
|
return 0
|
|
|
|
}
|
|
|
|
|
|
|
|
// Consume the search prefix
|
|
|
|
if len(child.prefix) > len(prefix) {
|
|
|
|
prefix = prefix[len(prefix):]
|
|
|
|
} else {
|
|
|
|
prefix = prefix[len(child.prefix):]
|
|
|
|
}
|
|
|
|
return t.deletePrefix(n, child, prefix)
|
|
|
|
}
|
|
|
|
|
|
|
|
func (n *node[T]) mergeChild() {
|
|
|
|
e := n.edges[0]
|
|
|
|
child := e.node
|
|
|
|
n.prefix = n.prefix + child.prefix
|
|
|
|
n.leaf = child.leaf
|
|
|
|
n.edges = child.edges
|
|
|
|
}
|
|
|
|
|
|
|
|
// Get is used to lookup a specific key, returning
|
|
|
|
// the value and if it was found
|
|
|
|
func (t *Tree[T]) Get(s string) (T, bool) {
|
|
|
|
var zeroVal T
|
|
|
|
n := t.root
|
|
|
|
search := s
|
|
|
|
for {
|
|
|
|
// Check for key exhaution
|
|
|
|
if len(search) == 0 {
|
|
|
|
if n.isLeaf() {
|
|
|
|
return n.leaf.val, true
|
|
|
|
}
|
|
|
|
break
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for an edge
|
|
|
|
n = n.getEdge(search[0])
|
|
|
|
if n == nil {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
|
|
|
|
// Consume the search prefix
|
|
|
|
if strings.HasPrefix(search, n.prefix) {
|
|
|
|
search = search[len(n.prefix):]
|
|
|
|
} else {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
}
|
|
|
|
return zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// LongestPrefix is like Get, but instead of an
|
|
|
|
// exact match, it will return the longest prefix match.
|
|
|
|
func (t *Tree[T]) LongestPrefix(s string) (string, T, bool) {
|
|
|
|
var zeroVal T
|
|
|
|
var last *leafNode[T]
|
|
|
|
n := t.root
|
|
|
|
search := s
|
|
|
|
for {
|
|
|
|
// Look for a leaf node
|
|
|
|
if n.isLeaf() {
|
|
|
|
last = n.leaf
|
|
|
|
}
|
|
|
|
|
|
|
|
// Check for key exhaution
|
|
|
|
if len(search) == 0 {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for an edge
|
|
|
|
n = n.getEdge(search[0])
|
|
|
|
if n == nil {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
|
|
|
|
// Consume the search prefix
|
|
|
|
if strings.HasPrefix(search, n.prefix) {
|
|
|
|
search = search[len(n.prefix):]
|
|
|
|
} else {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
}
|
|
|
|
if last != nil {
|
|
|
|
return last.key, last.val, true
|
|
|
|
}
|
|
|
|
return "", zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// Minimum is used to return the minimum value in the tree
|
|
|
|
func (t *Tree[T]) Minimum() (string, T, bool) {
|
|
|
|
var zeroVal T
|
|
|
|
n := t.root
|
|
|
|
for {
|
|
|
|
if n.isLeaf() {
|
|
|
|
return n.leaf.key, n.leaf.val, true
|
|
|
|
}
|
|
|
|
if len(n.edges) > 0 {
|
|
|
|
n = n.edges[0].node
|
|
|
|
} else {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
}
|
|
|
|
return "", zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// Maximum is used to return the maximum value in the tree
|
|
|
|
func (t *Tree[T]) Maximum() (string, T, bool) {
|
|
|
|
var zeroVal T
|
|
|
|
n := t.root
|
|
|
|
for {
|
|
|
|
if num := len(n.edges); num > 0 {
|
|
|
|
n = n.edges[num-1].node
|
|
|
|
continue
|
|
|
|
}
|
|
|
|
if n.isLeaf() {
|
|
|
|
return n.leaf.key, n.leaf.val, true
|
|
|
|
}
|
|
|
|
break
|
|
|
|
}
|
|
|
|
return "", zeroVal, false
|
|
|
|
}
|
|
|
|
|
|
|
|
// Walk is used to walk the tree
|
|
|
|
func (t *Tree[T]) Walk(fn WalkFn[T]) {
|
|
|
|
recursiveWalk(t.root, fn)
|
|
|
|
}
|
|
|
|
|
|
|
|
// WalkPrefix is used to walk the tree under a prefix
|
|
|
|
func (t *Tree[T]) WalkPrefix(prefix string, fn WalkFn[T]) {
|
|
|
|
n := t.root
|
|
|
|
search := prefix
|
|
|
|
for {
|
|
|
|
// Check for key exhaustion
|
|
|
|
if len(search) == 0 {
|
|
|
|
recursiveWalk(n, fn)
|
|
|
|
return
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for an edge
|
|
|
|
n = n.getEdge(search[0])
|
|
|
|
if n == nil {
|
|
|
|
return
|
|
|
|
}
|
|
|
|
|
|
|
|
// Consume the search prefix
|
|
|
|
if strings.HasPrefix(search, n.prefix) {
|
|
|
|
search = search[len(n.prefix):]
|
|
|
|
continue
|
|
|
|
}
|
|
|
|
if strings.HasPrefix(n.prefix, search) {
|
|
|
|
// Child may be under our search prefix
|
|
|
|
recursiveWalk(n, fn)
|
|
|
|
}
|
|
|
|
return
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
// WalkPath is used to walk the tree, but only visiting nodes
|
|
|
|
// from the root down to a given leaf. Where WalkPrefix walks
|
|
|
|
// all the entries *under* the given prefix, this walks the
|
|
|
|
// entries *above* the given prefix.
|
|
|
|
func (t *Tree[T]) WalkPath(path string, fn WalkFn[T]) {
|
|
|
|
n := t.root
|
|
|
|
search := path
|
|
|
|
for {
|
|
|
|
// Visit the leaf values if any
|
|
|
|
if n.leaf != nil && fn(n.leaf.key, n.leaf.val) {
|
|
|
|
return
|
|
|
|
}
|
|
|
|
|
|
|
|
// Check for key exhaution
|
|
|
|
if len(search) == 0 {
|
|
|
|
return
|
|
|
|
}
|
|
|
|
|
|
|
|
// Look for an edge
|
|
|
|
n = n.getEdge(search[0])
|
|
|
|
if n == nil {
|
|
|
|
return
|
|
|
|
}
|
|
|
|
|
|
|
|
// Consume the search prefix
|
|
|
|
if strings.HasPrefix(search, n.prefix) {
|
|
|
|
search = search[len(n.prefix):]
|
|
|
|
} else {
|
|
|
|
break
|
|
|
|
}
|
|
|
|
}
|
|
|
|
}
|
|
|
|
|
|
|
|
// recursiveWalk is used to do a pre-order walk of a node
|
|
|
|
// recursively. Returns true if the walk should be aborted
|
|
|
|
func recursiveWalk[T any](n *node[T], fn WalkFn[T]) bool {
|
|
|
|
// Visit the leaf values if any
|
|
|
|
if n.leaf != nil && fn(n.leaf.key, n.leaf.val) {
|
|
|
|
return true
|
|
|
|
}
|
|
|
|
|
|
|
|
// Recurse on the children
|
|
|
|
i := 0
|
|
|
|
k := len(n.edges) // keeps track of number of edges in previous iteration
|
|
|
|
for i < k {
|
|
|
|
e := n.edges[i]
|
|
|
|
if recursiveWalk(e.node, fn) {
|
|
|
|
return true
|
|
|
|
}
|
|
|
|
// It is a possibility that the WalkFn modified the node we are
|
|
|
|
// iterating on. If there are no more edges, mergeChild happened,
|
|
|
|
// so the last edge became the current node n, on which we'll
|
|
|
|
// iterate one last time.
|
|
|
|
if len(n.edges) == 0 {
|
|
|
|
return recursiveWalk(n, fn)
|
|
|
|
}
|
|
|
|
// If there are now less edges than in the previous iteration,
|
|
|
|
// then do not increment the loop index, since the current index
|
|
|
|
// points to a new edge. Otherwise, get to the next index.
|
|
|
|
if len(n.edges) >= k {
|
|
|
|
i++
|
|
|
|
}
|
|
|
|
k = len(n.edges)
|
|
|
|
}
|
|
|
|
return false
|
|
|
|
}
|
|
|
|
|
|
|
|
// ToMap is used to walk the tree and convert it into a map
|
|
|
|
func (t *Tree[T]) ToMap() map[string]T {
|
|
|
|
out := make(map[string]T, t.size)
|
|
|
|
t.Walk(func(k string, v T) bool {
|
|
|
|
out[k] = v
|
|
|
|
return false
|
|
|
|
})
|
|
|
|
return out
|
|
|
|
}
|