binary-search-tree: left() and right() return modifiable subtree

binary-search-tree の演習から始め、example.h を調べて、binary_tree::left()、binary_tree::right()、insert() に注目してください。それらの所有権と const 性を、issue に示されている現在のテストと比較してください。完了とは、例とテストが、木の不変条件を無効にできない const-safe な部分木 API で一致している状態です。

In the exercise "binary-search-tree" the methods left() and right() are hard to implement correctly. The example implementation itself is IMHO incorrect.

The tests look like this:

template<typename T>
using tree_ptr = typename std::unique_ptr<binary_tree::binary_tree<T>>;

template<typename T>
static void test_leaf(const tree_ptr<T> &tree, const T& data, bool has_left, bool has_right)

//...

test_leaf<uint32_t>(tested->left(), 2, false, false);

That forces implementations of binary_tree::left() (and binary_tree::right()) to return a tree_ptr reference or rvaue which points to a non-const subtree.
Remember: const std::unique_ptr<some_type> means that the unique_ptr itself is const, not the object it points to.

Somebody could take the example implementation (example.h) and write

auto tree = binary_tree::binary_tree<int>(100);
tree.insert(50);
tree.left()->insert(150);

and thus invalidate the invariant of tree.

I consider any implementation of a binary search tree that can be corrupted that way as faulty. I came up with three alternatives that avoid this issue and pass the current tests:

• binary_tree could have a member variable allow_insert that is true for the root and false for all subtrees
• binary_tree could have a parent pointer and insert() could check for each of its parents if the new value violates that parent's invariant
• left() and right() could copy the subtree and return a unique_ptr to that copy.

But IMHO none of those alternatives feels right.

The easiest solution would be if left() and right() could return const raw pointers. But one could argue that raw pointers result in unclear ownership.