问题描述
我正在尝试在 Rust 中实现二叉搜索树。 remove 方法使用了不安全的 Rust 代码,我想编写等效的安全 Rust 代码。
我遇到了问题,因为 self 是一个可变引用,当使用 match 语句解构它时,编译器会抱怨,因为我无法重新分配给它,因为它已经被借用了。
use std::cmp::Ordering;
#[derive(Debug)]
struct Node<T: Ord> {
key: T,left: Tree<T>,right: Tree<T>,}
#[derive(Debug)]
struct Tree<T: Ord>(Option<Box<Node<T>>>);
#[derive(Debug)]
pub struct BinarySearchTree<T: Ord> {
root: Tree<T>,}
impl<T: Ord> Node<T> {
fn new(value: T) -> Self {
Node {
key: value,left: Tree(None),right: Tree(None),}
}
}
impl<T: Ord> Tree<T> {
fn add(&mut self,value: T) {
let mut current = self;
while let Some(ref mut node) = current.0 {
match node.key.cmp(&value) {
Ordering::Less => current = &mut node.right,Ordering::Greater => current = &mut node.left,Ordering::Equal => current = &mut node.right,}
}
current.0 = Some(Box::new(Node::new(value)));
}
fn successor(&self,value: &T) -> Option<&T> {
let mut current = self.0.as_ref();
let mut successor = None;
while current.is_some() {
let node = current.unwrap();
if *value < node.key {
successor = current;
current = node.left.0.as_ref();
} else {
current = node.right.0.as_ref();
}
}
successor.map(|node| &node.key)
}
fn extract_min(&mut self) -> Option<T> {
let mut node = None;
if self.0.is_some() {
let mut current = self;
while current.0.as_ref().unwrap().left.0.is_some() {
current = &mut current.0.as_mut().unwrap().left;
}
let temp = current.0.take().unwrap();
node = Some(temp.key);
current.0 = temp.right.0;
}
node
}
fn extract_max(&mut self) -> Option<T> {
let mut node = None;
if self.0.is_some() {
let mut current = self;
while current.0.as_ref().unwrap().right.0.is_some() {
current = &mut current.0.as_mut().unwrap().right;
}
let temp = current.0.take().unwrap();
node = Some(temp.key);
current.0 = temp.left.0;
}
node
}
fn remove(&mut self,value: &T) {
let mut current: *mut Tree<T> = self;
unsafe {
while let Some(ref mut node) = (*current).0 {
match node.key.cmp(value) {
Ordering::Less => current = &mut node.right,Ordering::Equal => match (node.left.0.as_mut(),node.right.0.as_mut()) {
(None,None) => (*current).0 = None,(Some(_),None) => (*current).0 = node.left.0.take(),(None,Some(_)) => (*current).0 = node.right.0.take(),Some(_)) => {
(*current).0.as_mut().unwrap().key = node.right.extract_min().unwrap();
}
},}
}
}
}
}
impl<T: Ord> BinarySearchTree<T> {
pub fn new() -> Self {
BinarySearchTree { root: Tree(None) }
}
pub fn add(&mut self,value: T) {
self.root.add(value);
}
pub fn remove(&mut self,value: &T) {
self.root.remove(value);
}
pub fn successor(&self,value: &T) -> Option<&T> {
self.root.successor(value)
}
}
解决方法
steffahn 在 users.rust-lang.org 上解决了这个问题:
fn remove(&mut self,value: &T) {
let mut current = self;
while let Some(ref mut node) = current.0 {
match node.key.cmp(value) {
Ordering::Less => current = &mut current.0.as_mut().unwrap().right,Ordering::Greater => current = &mut current.0.as_mut().unwrap().left,Ordering::Equal => match (node.left.0.as_mut(),node.right.0.as_mut()) {
(None,None) => current.0 = None,(Some(_),None) => current.0 = node.left.0.take(),(None,Some(_)) => current.0 = node.right.0.take(),Some(_)) => {
current.0.as_mut().unwrap().key = node.right.extract_min().unwrap();
}
}
}
}
}
据作者说,
最重要的变化是在 node
和 Less
分支上重新跟踪 Greater
的路径(使用 unwrap
等)。这可能是编译器优化的零开销。