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oop - 为二叉树实现 IntoIterator

转载 作者:行者123 更新时间:2023-11-29 08:29:59 24 4
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我正在尝试构建一个二叉树并编写一个迭代器来遍历树中的值。在为我的树节点实现 IntoIterator 特性时,我遇到了生命周期问题

src\main.rs:43:6: 43:8 error: the lifetime parameter `'a` is not constrained by the impl trait, self type, or predicates [E0207]
src\main.rs:43 impl<'a, T: 'a> IntoIterator for Node<T> {

我知道我需要指定 NodeIterator 将与 Node 一样长,但我不确定如何表达这一点

use std::cmp::PartialOrd;
use std::boxed::Box;

struct Node<T: PartialOrd> {
value: T,
left: Option<Box<Node<T>>>,
right: Option<Box<Node<T>>>,
}

struct NodeIterator<'a, T: 'a + PartialOrd> {
current: &'a Node<T>,
parent: Option<&'a Node<T>>,
}

impl<T: PartialOrd> Node<T> {
pub fn insert(&mut self, value: T) {
...
}
}

impl<'a, T: 'a> IntoIterator for Node<T> { // line 43
type Item = T;
type IntoIter = NodeIterator<'a, T>;

fn into_iter(&self) -> Self::IntoIter {
NodeIterator::<'a> {
current: Some(&self),
parent: None
}
}
}

最佳答案

您遇到的特定错误是 'a应该出现在for的右边.否则,编译器怎么知道 a 是什么?是吗?

实现IntoIterator时您必须决定迭代器是消费容器,还是只生成对容器的引用。目前,您的设置不一致,错误消息指出了这一点。

在二叉树的情况下,您还必须考虑生成值的顺序:传统顺序是深度优先(产生排序序列)和广度优先(暴露树的“层” ).我将首先假设深度,因为它是最常见的。


让我们先解决消费迭代器的情况。从某种意义上说,它更简单,我们不必担心生命周期。

#![feature(box_patterns)]

struct Node<T: PartialOrd> {
value: T,
left: Option<Box<Node<T>>>,
right: Option<Box<Node<T>>>,
}

struct NodeIterator<T: PartialOrd> {
stack: Vec<Node<T>>,
next: Option<T>,
}

impl<T: PartialOrd> IntoIterator for Node<T> {
type Item = T;
type IntoIter = NodeIterator<T>;

fn into_iter(self) -> Self::IntoIter {
let mut stack = Vec::new();

let smallest = pop_smallest(self, &mut stack);

NodeIterator { stack: stack, next: Some(smallest) }
}
}

impl<T: PartialOrd> Iterator for NodeIterator<T> {
type Item = T;

fn next(&mut self) -> Option<T> {
if let Some(next) = self.next.take() {
return Some(next);
}

if let Some(Node { value, right, .. }) = self.stack.pop() {
if let Some(right) = right {
let box right = right;
self.stack.push(right);
}
return Some(value);
}

None
}
}

fn pop_smallest<T: PartialOrd>(node: Node<T>, stack: &mut Vec<Node<T>>) -> T {
let Node { value, left, right } = node;

if let Some(left) = left {
stack.push(Node { value: value, left: None, right: right });
let box left = left;
return pop_smallest(left, stack);
}

if let Some(right) = right {
let box right = right;
stack.push(right);
}

value
}

fn main() {
let root = Node {
value: 3,
left: Some(Box::new(Node { value: 2, left: None, right: None })),
right: Some(Box::new(Node { value: 4, left: None, right: None }))
};

for t in root {
println!("{}", t);
}
}

现在,我们可以通过添加适当的引用来“轻松”地使其适应非消费情况:

struct RefNodeIterator<'a, T: PartialOrd + 'a> {
stack: Vec<&'a Node<T>>,
next: Option<&'a T>,
}

impl<'a, T: PartialOrd + 'a> IntoIterator for &'a Node<T> {
type Item = &'a T;
type IntoIter = RefNodeIterator<'a, T>;

fn into_iter(self) -> Self::IntoIter {
let mut stack = Vec::new();

let smallest = pop_smallest_ref(self, &mut stack);

RefNodeIterator { stack: stack, next: Some(smallest) }
}
}

impl<'a, T: PartialOrd + 'a> Iterator for RefNodeIterator<'a, T> {
type Item = &'a T;

fn next(&mut self) -> Option<&'a T> {
if let Some(next) = self.next.take() {
return Some(next);
}

if let Some(node) = self.stack.pop() {
if let Some(ref right) = node.right {
self.stack.push(right);
}
return Some(&node.value);
}

None
}
}

fn pop_smallest_ref<'a, T>(node: &'a Node<T>, stack: &mut Vec<&'a Node<T>>) -> &'a T
where
T: PartialOrd + 'a
{
if let Some(ref left) = node.left {
stack.push(node);
return pop_smallest_ref(left, stack);
}

if let Some(ref right) = node.right {
stack.push(right);
}

&node.value
}

里面有很多东西要打开;所以花点时间消化它。具体来说:

  • 使用refSome(ref right) = node.right是因为我不想消费node.right , 只获取Option里面的引用;编译器会提示我不能在没有它的情况下移出借来的对象(所以我只是听从了提示),
  • stack.push(right) , right: &'a Box<Node<T>>然而stack: Vec<&'a Node<T>> ;这就是Deref的魔力: Box<T>工具 Deref<T>因此编译器会根据需要自动转换引用。

注意:我没有按原样编写这段代码;相反,我只是将前几个引用放在我期望的位置(例如 Iterator 的返回类型),然后让编译器指导我。

关于oop - 为二叉树实现 IntoIterator,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43833588/

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