If else assignment "move out of borrowed content" - if-statement

Question

This is a snippet of Rust code which performs an insertion on a common binary search tree. I want the reference to the branch on which I'll perform the insertion.
struct Node {
value: i32,
left: Option<Box<Node>>,
right: Option<Box<Node>>,
}
impl Node {
fn insert(&mut self, elem: i32) {
let ref mut a = if elem < self.value {
self.left
} else {
self.right
};
// ...
}
}
This code is invalid. After tweaking it, I came to understand when I perform a if/else statement, Rust moves the content before assigning it. This solution works, but it's really ugly...
let a = if elem < self.value {
let ref mut b = self.left;
b
} else {
let ref mut b = self.right;
b
};
Is there a way to handle the thing without recurring to a Box? I could use a pointer, but it really looks like overkill.
After commenting, this is the entire code
fn insert(&mut self, elem: i32) {
let target = if elem < self.value {
&mut self.left
} else {
&mut self.right
};
match target.as_mut() {
None => {}
Some(ref mut node) => {
node.insert(elem);
return;
}
}
mem::replace(&mut *target, Some(Box::new(Node::new(elem))));
}

Answer 1

As you observed, in your original attempt the "then" and "else" branches both attempt to move the left or right box out of your &mut self borrowed content, and this is not allowed.
let ref mut a = if elem < self.value {
self.left
} else {
self.right
};
Your second attempt compiles but is ugly, as you said. This code avoids moving out of the borrowed content by referring to the left and right box by ref mut mutable reference.
let a = if elem < self.value {
let ref mut b = self.left;
b
} else {
let ref mut b = self.right;
b
};
A cleaner way to refer to something by mutable reference is with the &mut operator. The if-statement should be:
let a = if elem < self.value {
&mut self.left
} else {
&mut self.right
};