rust - 如何实现对二叉搜索树右边缘值的可变引用的迭代器？

Question

我在 Rust 中实现了一个简单的二叉搜索树(遵循 CIS 198，这很棒)，为了学习，我正在做的迭代器只是通过右边缘运行。

我无法实现提供可变引用的迭代器。我尝试了很多方法，但没有一个被 Rust 编译器接受。我需要帮助的代码是下面的 (while I made a gist with the complete code here):

#[derive(Debug)]
pub struct Tree<T>(Option<Box<Node<T>>>);

#[derive(Debug)]
pub struct Node<T> {
    elem: T,
    left: Tree<T>,
    right: Tree<T>,
}

// MUTABLE BORROW STRUCT
pub struct IterMut<'a, T: 'a> {
    next: &'a mut Tree<T>,
}

// MUTABLE BORROW NEXT (I'M STUCK HERE, NOTHING WORKS)
impl<'a, T> Iterator for IterMut<'a, T> {
    type Item = &'a mut T;
    fn next(&mut self) -> Option<Self::Item> {
        // 1 try: cannot infer lifetime
        self.next.0.as_mut().map(|node| {
            self.next = &mut node.right;
            &mut node.elem
        })

        // 2 try: node.right, node.elem does not live long enough
        self.next.0.take().map(|node| {
            self.next = &mut node.right;
            &mut node.elem
        })
    }
}

Answer 1

您需要更改字段类型 IterMut::next至 Option<&'a mut Node<T>> :

pub struct IterMut<'a, T: 'a> {
    next: Option<&'a mut Node<T>>,
}

impl<'a, T> Iterator for IterMut<'a, T> {
    type Item = &'a mut T;
    fn next(&mut self) -> Option<Self::Item> {
        self.next.take().map(|node| {
            self.next = node.right.0.as_mut().map(|node| &mut **node);
            &mut node.elem
        })

    }
}

您可以找到有关递归数据结构的可变迭代器实现的更多有用信息 in the IterMut chapter of Learning Rust With Entirely Too Many Linked Lists .