//! This data structure is a generic tree which can contain any data. That data itself need to keep
//! track of its own tree structure.
//!
//! This surely already exists. I am created it to test my own ability to do things in Rust.

use std::{
    cell::{Ref, RefCell},
    collections::VecDeque,
    rc::Rc,
};

#[derive(Debug)]
pub enum Tree<T> {
    Empty,
    Root(Node<T>),
}

impl<T> Default for Tree<T> {
    fn default() -> Self {
        Tree::Empty
    }
}

impl<T> Tree<T> {
    pub fn set_value(&mut self, value: T) {
        *self = Tree::Root(Node::new(value));
    }

    /// Use a breadth-first-search pattern to find a node, returning the node if found.
    pub fn find_bfs<'a, F>(&'a self, op: F) -> Option<Node<T>>
    where
        F: FnOnce(&T) -> bool + Copy,
    {
        let mut queue: VecDeque<Node<T>> = match self {
            Tree::Empty => VecDeque::new(),
            Tree::Root(node) => {
                let mut queue = VecDeque::new();
                queue.push_back(node.clone());
                queue
            }
        };

        while let Some(node) = queue.pop_front() {
            if op(&node.value()) {
                return Some(node.clone());
            }

            for child in node.children().iter() {
                queue.push_back(child.clone())
            }
        }
        None
    }
}

// By using the Rc<RefCell> container here, I'm able to make Node easily clonable while still
// having the contents be shared. This means that I can change the tree structure without having to
// make the visible objects mutable.
//
// This feels like cheating the type system.
//
// However, since I've moved the RefCell inside of the node, I can borrow the node multiple times
// in a traversal function and I can make changes to nodes that I find.
#[derive(Debug)]
pub struct Node<T>(Rc<RefCell<Node_<T>>>);

impl<T> Clone for Node<T> {
    fn clone(&self) -> Self {
        Node(Rc::clone(&self.0))
    }
}

#[derive(Debug)]
struct Node_<T> {
    value: T,
    children: Vec<Node<T>>,
}

impl<T> Node<T> {
    pub fn new(value: T) -> Self {
        Self(Rc::new(RefCell::new(Node_ {
            value,
            children: vec![],
        })))
    }

    // I am copying the value because I don't have a mechanism for keeping the borrow ref active.
    // My next step is to figure out how to keep the borrow ref active so that I don't have to
    // clone the item. Then I could drop the Clone constraint.
    pub fn value<'a>(&'a self) -> Ref<T> {
        Ref::map(self.0.borrow(), |v| &v.value)
    }

    pub fn children<'a>(&'a self) -> Ref<Vec<Node<T>>> {
        Ref::map(self.0.borrow(), |v| &v.children)
    }

    pub fn add_child_node(&self, child: Node<T>) {
        self.0.borrow_mut().children.push(child)
    }

    pub fn add_child_value(&self, value: T) {
        self.0.borrow_mut().children.push(Node::new(value))
    }
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn can_find_node_in_tree() {
        let mut tree = Tree::default();
        tree.set_value(15);
        assert!(tree.find_bfs(|val| *val == 15).is_some());
        assert!(tree.find_bfs(|val| *val == 16).is_none());

        let node = tree.find_bfs(|val| *val == 15).unwrap();
        node.add_child_value(20);

        assert!(tree.find_bfs(|val| *val == 20).is_some());
    }

    #[test]
    fn node_can_add_children() {
        let n = Node::new(15);
        n.add_child_value(20);

        assert_eq!(*n.value(), 15);
        // assert_eq!(n.children(), vec![Rc::new(RefCell::new(Node::new(20)))]);
    }
}