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);
}

playground

但是，我收到以下编译错误：

error[E0502]: cannot borrow `*self` as immutable because `self.left` is also borrowed as mutable
  --> src/main.rs:24:42
   |
23 |             Tree::Node {ref mut left, ..} => {
   |                         ------------ mutable borrow occurs here
24 |                 **left = Tree::new(data, self);
   |                                          ^^^^ immutable borrow occurs here
25 |             }
26 |         }
   |         - mutable borrow ends here

在安全的Rust中执行此操作的范例方法是什么？具体来说，当我插入一个新节点作为现有节点的叶子时，我不想为它重新分配空间。已经为Leaf分配了空间，我想简单地用新节点覆盖它。

使用父指针创建Rust树的范例方法是什么？

0 个答案: