rust - 深度优先树搜索期间的多个可变借用

Question

如何重构这个进行深度优先搜索并返回匹配节点父节点的函数？

我知道这个问题的变体经常出现(例如 Multiple mutable borrows when generating a tree structure with a recursive function in Rust ， Mut borrow not ending where expected )，但我仍然不知道如何修改它才能工作。我已经尝试使用切片、std::mem::drop 和添加生命周期参数 where 'a: 'b 的变体，但我仍然没有想出不写它条件可变地在一个分支中借用一个变量，然后在另一个分支中使用该变量。

#[derive(Clone, Debug)]
struct TreeNode {
    id: i32,
    children: Vec<TreeNode>,
}

// Returns a mutable reference to the parent of the node that matches the given id.
fn find_parent_mut<'a>(root: &'a mut TreeNode, id: i32) -> Option<&'a mut TreeNode> {
    for child in root.children.iter_mut() {
        if child.id == id {
            return Some(root);
        } else {
            let descendent_result = find_parent_mut(child, id);
            if descendent_result.is_some() {
                return descendent_result;
            }
        }
    }
    None
}

fn main() {
    let mut tree = TreeNode {
        id: 1,
        children: vec![TreeNode {
                           id: 2,
                           children: vec![TreeNode {
                                              id: 3,
                                              children: vec![],
                                          }],
                       }],
    };
    let a: Option<&mut TreeNode> = find_parent_mut(&mut tree, 3);
    assert_eq!(a.unwrap().id, 2);
}

error[E0499]: cannot borrow `*root` as mutable more than once at a time
  --> src/main.rs:11:25
   |
9  |     for child in root.children.iter_mut() {
   |                  ------------- first mutable borrow occurs here
10 |         if child.id == id {
11 |             return Some(root);
   |                         ^^^^ second mutable borrow occurs here
...
20 | }
   | - first borrow ends here

这是@huon 的建议和持续的编译器错误:

fn find_parent_mut<'a>(root: &'a mut TreeNode, id: i32) -> Option<&'a mut TreeNode> {
    for child in root.children {
        if child.id == id {
            return Some(root);
        }
    }
    for i in 0..root.children.len() {
        let child: &'a mut TreeNode = &mut root.children[i];
        let descendent_result = find_parent_mut(child, id);
        if descendent_result.is_some() {
            return descendent_result;
        }
    }
    None
}

error[E0507]: cannot move out of borrowed content
 --> src/main.rs:9:18
  |
9 |     for child in root.children {
  |                  ^^^^ cannot move out of borrowed content

error[E0499]: cannot borrow `root.children` as mutable more than once at a time
  --> src/main.rs:15:44
   |
15 |         let child: &'a mut TreeNode = &mut root.children[i];
   |                                            ^^^^^^^^^^^^^
   |                                            |
   |                                            second mutable borrow occurs here
   |                                            first mutable borrow occurs here
...
22 | }
   | - first borrow ends here

Answer 1

我设法让它以这种方式工作:

fn find_parent_mut<'a>(root: &'a mut TreeNode, id: i32)
        -> Option<&'a mut TreeNode> {
    if root.children.iter().any(|child| {child.id == id}) {
        return Some(root); 
    }
    for child in &mut root.children {
        match find_parent_mut(child, id) {
            Some(result) => return Some(result),
            None => {}
        }
    }
    None
}

第一次在第二次尝试中，您编写了 for child in root.children 而不是 for child in &mut root.children(注意缺少的 &mut)，这导致 root.children 被循环消耗，而不是仅仅迭代，因此 cannot move out of borrowed content 错误。

我还使用 any(..) 以一种更像迭代器的方式折叠它功能。

对于第二个循环，我不太确定发生了什么，显然将引用绑定(bind)到变量使借用检查器感到困惑。我删除了任何临时变量，现在可以编译了。