Multiple mutable borrows during a depth-first tree

问题:

How would one restructure this function that does a depth-first search and returns the parent of the matching node?

I know that variations of this problem have come up very often (e.g. Multiple mutable borrows when generating a tree structure with a recursive function in Rust, Mut borrow not ending where expected), but I still can’t figure out how to modify it to work. I have tried variations using slices, std::mem::drop, and adding lifetime parameters where 'a: 'b but I still haven’t figured out write it without conditionally mutably borrowing a variable in one branch and then using the variable in another branch.

#[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

Here it is with @huon’s suggestions and continued compiler errors:

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

回答1:

I managed to have it working this way:

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
}

First in you second attempt, you wrote for child in root.children instead of the for child in &mut root.children (note the missing &mut),which caused root.children to be consumed by the loop instead of just iterated over, hence the cannot move out of borrowed content error.

I also folded it in a more iterator-ich way, using the any(..) function.

For the second loop, I'm not exactly sure what was going on, by apparently binding the references to variables was confusing the borrow-checker. I removed any temporary variable, and now it compiles.

回答2:

One can do this in a single pass by doing the search, recording some data from it, and then computing the return value efficiently outside the loop:

let mut found = Err(());
for (i, child) in root.children.iter_mut().enumerate() {
    if child.id == id {
        found = Ok(None);
        break;
    }  else find_parent_mut(child, id).is_some() {
        found = Ok(Some(i));
        break;
    }
}
match found {
    Ok(Some(i)) => Some(&mut root.children[i]),
    Ok(None) => Some(root),
    Err(()) => None,
}

This avoids the issues caused by conditionally returning mutable variables (which is the problems you and my answer below were meeting) by completely avoiding returns inside the inner loop.

---

Old/incorrect answer:

I'm unable to test my suggestions right now, but I believe the best way to solve this is to return root outside the loop.

for child in &mut root.children {
    if child.id == id {
        found = true;
        break
    } ...
}
if found {
    Some(root)
} else {
    None
}

This (hopefully) ensures that root isn't borrowed via children when it is being manipulated.

However I suspect the early return inside the main loop may interfere, in which case one might just have to fall back to integers and indexing:

for i in 0..root.children.len() {
    if root.children[i].id == id {
        return Some(root)
    ...
}