使用父指针创建 Rust 树的典型方法是什么？

Question

我需要定义一个二叉搜索树，其中每个节点都可以访问父节点：

enum Tree<'a> {
    Leaf,
    Node {
        left: Box<Tree<'a>>,
        right: Box<Tree<'a>>,
        parent: &'a Tree<'a>,
        data: u64,
    }
}

impl <'a> Tree<'a> {
    pub fn new(data: u64, parent: &'a Tree) -> Tree<'a> {
        Tree::Node {
            left: Box::new(Tree::Leaf),
            right: Box::new(Tree::Leaf),
            parent,
            data
        }
    }
    pub fn insert_at_left_leaf(&'a mut self, data: u64) {
        match *self {
            Tree::Leaf => panic!("Leaf has no children"),
            Tree::Node {ref mut left, ..} => {
                **left = Tree::new(data, self);
            }
        }
    }
}

fn main() {
    let parent = Tree::Leaf;
    let mut t = Tree::Node {
        left: Box::new(Tree::Leaf),
        right: Box::new(Tree::Leaf),
        parent: &parent,
        data: 1u64
    };
    t.insert_at_left_leaf(2);
}

操场

但是，我收到以下编译错误：

enum Tree<'a> {
    Leaf,
    Node {
        left: Box<Tree<'a>>,
        right: Box<Tree<'a>>,
        parent: &'a Tree<'a>,
        data: u64,
    }
}

impl <'a> Tree<'a> {
    pub fn new(data: u64, parent: &'a Tree) -> Tree<'a> {
        Tree::Node {
            left: Box::new(Tree::Leaf),
            right: Box::new(Tree::Leaf),
            parent,
            data
        }
    }
    pub fn insert_at_left_leaf(&'a mut self, data: u64) {
        match *self {
            Tree::Leaf => panic!("Leaf has no children"),
            Tree::Node {ref mut left, ..} => {
                **left = Tree::new(data, self);
            }
        }
    }
}

fn main() {
    let parent = Tree::Leaf;
    let mut t = Tree::Node {
        left: Box::new(Tree::Leaf),
        right: Box::new(Tree::Leaf),
        parent: &parent,
        data: 1u64
    };
    t.insert_at_left_leaf(2);
}

在安全的 Rust 中这样做的典型方法是什么？具体来说，当我插入一个新节点作为现有节点的叶子时，我不想为其重新分配空间。已经为 分配了空间Leaf，我想简单地用新节点覆盖它。