Rust tries to ensure memory safety by forbidding you from doing things that might potentially be unsafe. Since this analysis is performed at compile-time, the compiler can only reason about a subset of manipulations that are known to be safe.
In Rust, you could easily store either a reference to the parent (by borrowing the parent, thus preventing mutation) or the list of child nodes (by owning them, which gives you more freedom), but not both (without using unsafe). This is especially problematic for your implementation of addNode, which requires mutable access to the given node's parent. However, if you store the parent pointer as a mutable reference, then, since only a single mutable reference to a particular object may be usable at a time, the only way to access the parent would be through a child node, and you'd only be able to have a single child node, otherwise you'd have two mutable references to the same parent node.
If you want to avoid unsafe code, there are many alternatives, but they'll all require some sacrifices.
The easiest solution is to simply remove the parent field. We can define a separate data structure to remember the parent of a node while we traverse a tree, rather than storing it in the node itself.
First, let's define Node:
#[derive(Debug)]
struct Node<T> {
data: T,
children: Vec<Node<T>>,
}
impl<T> Node<T> {
fn new(data: T) -> Node<T> {
Node { data: data, children: vec![] }
}
fn add_child(&mut self, child: Node<T>) {
self.children.push(child);
}
}
(I added a data field because a tree isn't super useful without data at the nodes!)
Let's now define another struct to track the parent as we navigate:
#[derive(Debug)]
struct NavigableNode<'a, T: 'a> {
node: &'a Node<T>,
parent: Option<&'a NavigableNode<'a, T>>,
}
impl<'a, T> NavigableNode<'a, T> {
fn child(&self, index: usize) -> NavigableNode<T> {
NavigableNode {
node: &self.node.children[index],
parent: Some(self)
}
}
}
impl<T> Node<T> {
fn navigate<'a>(&'a self) -> NavigableNode<T> {
NavigableNode { node: self, parent: None }
}
}
This solution works fine if you don't need to mutate the tree as you navigate it and you can keep the parent NavigableNode objects around (which works fine for a recursive algorithm, but doesn't work too well if you want to store a NavigableNode in some other data structure and keep it around). The second restriction can be alleviated by using something other than a borrowed pointer to store the parent; if you want maximum genericity, you can use the Borrow trait to allow direct values, borrowed pointers, Boxes, Rc's, etc.
Now, let's talk about zippers. In functional programming, zippers are used to "focus" on a particular element of a data structure (list, tree, map, etc.) so that accessing that element takes constant time, while still retaining all the data of that data structure. If you need to navigate your tree and mutate it during the navigation, while retaining the ability to navigate up the tree, then you could turn a tree into a zipper and perform the modifications through the zipper.
Here's how we could implement a zipper for the Node defined above:
#[derive(Debug)]
struct NodeZipper<T> {
node: Node<T>,
parent: Option<Box<NodeZipper<T>>>,
index_in_parent: usize,
}
impl<T> NodeZipper<T> {
fn child(mut self, index: usize) -> NodeZipper<T> {
// Remove the specified child from the node's children.
// A NodeZipper shouldn't let its users inspect its parent,
// since we mutate the parents
// to move the focused nodes out of their list of children.
// We use swap_remove() for efficiency.
let child = self.node.children.swap_remove(index);
// Return a new NodeZipper focused on the specified child.
NodeZipper {
node: child,
parent: Some(Box::new(self)),
index_in_parent: index,
}
}
fn parent(self) -> NodeZipper<T> {
// Destructure this NodeZipper
let NodeZipper { node, parent, index_in_parent } = self;
// Destructure the parent NodeZipper
let NodeZipper {
node: mut parent_node,
parent: parent_parent,
index_in_parent: parent_index_in_parent,
} = *parent.unwrap();
// Insert the node of this NodeZipper back in its parent.
// Since we used swap_remove() to remove the child,
// we need to do the opposite of that.
parent_node.children.push(node);
let len = parent_node.children.len();
parent_node.children.swap(index_in_parent, len - 1);
// Return a new NodeZipper focused on the parent.
NodeZipper {
node: parent_node,
parent: parent_parent,
index_in_parent: parent_index_in_parent,
}
}
fn finish(mut self) -> Node<T> {
while let Some(_) = self.parent {
self = self.parent();
}
self.node
}
}
impl<T> Node<T> {
fn zipper(self) -> NodeZipper<T> {
NodeZipper { node: self, parent: None, index_in_parent: 0 }
}
}
To use this zipper, you need to have ownership of the root node of the tree. By taking ownership of the nodes, the zipper can move things around in order to avoid copying or cloning nodes. When we move a zipper, we actually drop the old zipper and create a new one (though we could also do it by mutating self, but I thought it was clearer that way, plus it lets you chain method calls).
If the above options are not satisfactory, and you must absolutely store the parent of a node in a node, then the next best option is to use Rc<RefCell<Node<T>>> to refer to the parent and Weak<RefCell<Node<T>>> to the children. Rc enables shared ownership, but adds overhead to perform reference counting at runtime. RefCell enables interior mutability, but adds overhead to keep track of the active borrows at runtime. Weak is like Rc, but it doesn't increment the reference count; this is used to break reference cycles, which would prevent the reference count from dropping to zero, causing a memory leak. See DK.'s answer for an implementation using Rc, Weak and RefCell.
std::vector<std::unique_ptr<Node>> children;? Also, why is the destructorvirtual? Is this supposed to be an interface? – Hokanson