練習：二元樹

二元樹是一種樹狀資料結構，其中每個節點都有左右兩個子節點。我們會建立每個節點都儲存一個值的樹狀結構。以指定節點 N 來說，N 左側子樹狀結構中的所有節點都包含較小的值，而 N 右側子樹狀結構中的所有節點都含有較大的值。

請實作以下型別，讓指定的測試通過。

加分題：在二元數上實作疊代器，依序傳回值。

/// A node in the binary tree.
#[derive(Debug)]
struct Node<T: Ord> {
    value: T,
    left: Subtree<T>,
    right: Subtree<T>,
}

/// A possibly-empty subtree.
#[derive(Debug)]
struct Subtree<T: Ord>(Option<Box<Node<T>>>);

/// A container storing a set of values, using a binary tree.
///
/// If the same value is added multiple times, it is only stored once.
#[derive(Debug)]
pub struct BinaryTree<T: Ord> {
    root: Subtree<T>,
}

// Implement `new`, `insert`, `len`, and `has`.

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

    #[test]
    fn len() {
        let mut tree = BinaryTree::new();
        assert_eq!(tree.len(), 0);
        tree.insert(2);
        assert_eq!(tree.len(), 1);
        tree.insert(1);
        assert_eq!(tree.len(), 2);
        tree.insert(2); // not a unique item
        assert_eq!(tree.len(), 2);
    }

    #[test]
    fn has() {
        let mut tree = BinaryTree::new();
        fn check_has(tree: &BinaryTree<i32>, exp: &[bool]) {
            let got: Vec<bool> =
                (0..exp.len()).map(|i| tree.has(&(i as i32))).collect();
            assert_eq!(&got, exp);
        }

        check_has(&tree, &[false, false, false, false, false]);
        tree.insert(0);
        check_has(&tree, &[true, false, false, false, false]);
        tree.insert(4);
        check_has(&tree, &[true, false, false, false, true]);
        tree.insert(4);
        check_has(&tree, &[true, false, false, false, true]);
        tree.insert(3);
        check_has(&tree, &[true, false, false, true, true]);
    }

    #[test]
    fn unbalanced() {
        let mut tree = BinaryTree::new();
        for i in 0..100 {
            tree.insert(i);
        }
        assert_eq!(tree.len(), 100);
        assert!(tree.has(&50));
    }
}