Multiple mutable borrows during a depth-first tree search - rust

Question

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

Answer 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.

Answer 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)
...
}