[英]Implement IntoIterator for binary tree
我试图建立一个二叉树并编写一个迭代器以遍历树中的值。 当为我的树节点实现IntoIterator特性时,我遇到了生存期问题
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> {
我了解我需要指定NodeIterator的生存时间与Node一样长,但我不确定如何表达
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
}
}
}
您遇到的特定错误是'a
应该出现在for
的右边。 否则,编译器如何知道a
是什么?
在实现IntoIterator
您必须决定迭代器是否会使用容器,或者是否只是在容器中产生引用。 目前,您的设置不一致,并且错误消息指出了该错误。
对于二叉树,您还必须考虑要按哪个顺序生成值:传统顺序是深度优先(产生排序的序列)和宽度优先(暴露树的“层”)。 我首先假定深度,因为它是最常见的深度。
让我们首先解决一个消耗迭代器的情况。 从某种意义上讲,我们不必担心生命周期,这很简单。
#![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”使它适应非使用情况:
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
}
那里有很多东西可以解开。 所以花些时间来消化它。 特别:
Some(ref right) = node.right
使用ref
是因为我不想消耗node.right
,而只是在Option
内部获取引用; 编译器会抱怨说,没有它,我将无法移出借用的对象(因此,我只关注这些投诉), stack.push(right)
, right: &'a Box<Node<T>>
,但仍在stack: Vec<&'a Node<T>>
; 这就是Deref
的魔力: Box<T>
实现Deref<T>
因此编译器会根据需要自动转换引用。 注意:我没有按原样编写此代码; 相反,我只是将前几个引用放在期望的位置(例如Iterator
的返回类型),然后让编译器引导我。
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