You are an expert in writing code in many different languages. Your goal is to perform code completion for the following code, keeping in mind the rest of the code and the file meta data.
    Metadata details:
    filename: 0543-diameter-of-binary-tree.rs
    path of file: ./repos/NeetCode-Solutions/07_Trees/03_Diameter_of_Binary_Tree
    
    the code of the file until where you have to start completion: /*
Problem: LeetCode 543 - Diameter of Binary Tree

Key Idea:
The key idea is to use depth-first traversal. For each node, calculate the sum of the depths of its left and right subtrees. The diameter will be the maximum of these sums or the maximum diameter found in any subtree.

Approach:
1. Create a recursive function 'diameter_of_binary_tree' that takes a reference to the root of the binary tree and returns a tuple (depth, diameter).
2. If the root is None, return (0, 0) (base case).
3. Recursively calculate the depth and diameter of the left subtree by calling 'diameter_of_binary_tree' on the left child.
4. Recursively calculate the depth and diameter of the right subtree by calling 'diameter_of_binary_tree' on the right child.
5. Calculate the diameter as the maximum of
    
    pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        fn diameter_of_binary_tree(
            root: &Option<Rc<RefCell<TreeNode>>>,
            max_diameter: &mut i32,
        ) -> (i32, i32) {
            match root {
                None => (0, 0),
                Some(node) => {
                    let node = node.borrow();
                    let (left_depth, left_diameter) =
                        diameter_of_binary_tree(&node.left, max_diameter);
                    let (right_depth, right_diameter) =
                        diameter_of_binary_tree(&node.right, max_diameter);

                    let depth = left_depth.max(right_depth) + 1;
                    let diameter = left_diameter
                        .max(right_diameter)
                        .max(left_depth + right_depth);

                    *max_diameter = (*max_diameter).max(diameter);

                    (depth, diameter)
                }
            }
        }

        let mut max_diameter = 0;
        let (_, diameter) = diameter_of_binary_tree(&root, &mut max_diameter);
        diameter
    }