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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 特征时，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

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
         }
     }
 }
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1 回答 1

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您得到的特定错误'a应该出现在for. 否则，编译器怎么知道a是什么？

在实现时IntoIterator，您必须决定迭代器是否会使用容器，或者它是否只会产生对它的引用。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，您还必须考虑要以哪种顺序生成值：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。

让我们首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“轻松”地将其调整为非消耗案例：

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
}

里面有很多东西要解开；所以花点时间消化它。具体来说：

  • 使用refinSome(ref right) = node.right是因为我不想消费node.right，只是为了获取里面的引用Option；编译器会抱怨说没有它我不能搬出借来的对象（所以我只是听从抱怨），
  • 在stack.push(right)，right: &'a Box<Node<T>>然而stack: Vec<&'a Node<T>>；这是Deref:的Box<T>魔力，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我期望它们的位置（例如 的返回类型Iterator），然后让编译器指导我。

于 2017-05-07T17:13:50.017 回答