I am new to C++ and am attempting to write the following binary search tree using the iterator pattern (and std::optional in C++17):

#include <memory>
#include <optional>
#include <stack>

template <typename T>
class BinarySearchTree {
  BinarySearchTree(T key)
  : key_(key),
    right_(nullptr) {}

  BinarySearchTree(const BinarySearchTree& bst) {
    key_ = bst.key_;
    left_ = std::make_unique<BinarySearchTree>(*bst.left_);
    right_ = std::make_unique<BinarySearchTree>(*bst.right_);

  void insert(T key);

  T key() { return key_; }
  std::unique_ptr<BinarySearchTree> left() { return std::make_unique<BinarySearchTree>(*left_); }
  std::unique_ptr<BinarySearchTree> right() { return std::make_unique<BinarySearchTree>(*right_); }

  T key_;
  std::unique_ptr<BinarySearchTree> left_;
  std::unique_ptr<BinarySearchTree> right_;

template <typename T>
void BinarySearchTree<T>::insert(T key) {
  auto insert = [key](auto& node) {
    if (node) {
    } else {
      node = std::make_unique<BinarySearchTree>(key);

  if (key <= key_) {
  } else {

template <typename T>
class BinarySearchTreeIterator {
  BinarySearchTreeIterator(BinarySearchTree<T>& bst, bool forward)
  : bst_(std::make_unique<BinarySearchTree<T>>(bst)),
    forward_(forward) {}

  std::optional<T> operator*() { return current_; }

  void operator++() {
    while (bst_ || !stack_.empty()) {
      if (bst_) {
        bst_ = std::make_unique<BinarySearchTree<T>>(*(forward_ ? bst_->left() : bst_->right()));
      } else {
        bst_ = std::make_unique<BinarySearchTree<T>>(*(stack_.top()));
        current_ = bst_->key();
        bst_ = std::make_unique<BinarySearchTree<T>>(*(forward_ ? bst_->right() : bst_->left()));

  std::unique_ptr<BinarySearchTree<T>> bst_;
  bool forward_;
  std::stack<std::unique_ptr<BinarySearchTree<T>>> stack_;
  std::optional<T> current_;

I am not sure if my usage of std::unique_ptr is correct or idiomatic.


1 Answer 1



  • Since the tree isn't balanced, worst case insertion complexity is \$\mathcal{O}(n)\$ (and not \$\mathcal{O}(\log n)\$ as likely intended). (Same for lookup, if such a method were to be added).
  • Calling left() or right() will create a copy of the whole left/right subtree. This is very likely not intended.
  • This class could profit a lot from a move constructor and a move assignment operator overload. (And while we're at it, add a copy assignment operator overload and a destructor to complete the rule of five.)
  • Is it really necessary to name the lambda the same as the enclosing function in BinarySearchTree::insert(T key)?
  • Also of note, only types that can be copied can be inserted into the BinarySearchTree. This means it is currently not possibly to store a move-only type (like std::unique_ptr) in the tree. (This capacity could be added with an overload accepting a T&&.
  • BinarySearchTree::insert(T key) creates unnecessary copies of key for each recursive call. Consider taking a const T& instead.`
  • left_ and right_ are often dereferenced without checking them for nullptr beforehand! This results in undefined behavior if they are nullptr!


  • This iterator doesn't provide any iterator_traits or an end iterator. This means using it with standard library algorithms (or even a range based for loop) is not possible. More info on iterators and iterator_traits.
  • There are lots and lots of unnecessary copies (basically every time std::make_unique or bst_->left() or bst_->right() are called).
  • current_ will always have a value with this implementation, so making it std::optional<T> is misleading. Also, any modifications done to the value returned by operator*() won't be propagated to the original BinarySearchTree entry (as it returns an independent copy and not a reference).
  • This iterator traverses the tree in-order. Since there are also other traversal methods, like pre-order, post-order or level-order, it would be nice to indicate that in the name.

std::unique_ptr usage

Normally, a std::unique_ptr is used to express "I own this". So a signature like std::unique_ptr<BinarySearchTree<T>> BinarySearchTree::left() says "Call me, and you become the new owner of whatever I return".

Is this necessary in this case? No! After all, the caller just wants to access the existing object, and not acquire a new one.

So, how can this be expressed?

There are two variants:

  1. Return a non-owning BinarySearchTree<T>*.


    • Very lightweight.
    • Very simple (can still use std::unique_ptr for internal storage, so resources get cleaned up properly).


    • If the BinarySearchTree<T> gets destroyed, but someone still has the pointer stored, the pointer is dangling.
    • Someone could prematurely call delete on the pointer (so the internal std::unique_ptr would be dangling).
  2. Return a std::shared_ptr<BinarySearchTree<T>>.


    • As long as the std::shared_ptr exists, the object will be valid.


    • Might keep the tree (or subtrees) alive for far longer than intended.
    • Has a slight overhead.
    • Requires using std::shared_ptr for internal storage.
    • Might leak in case of cycles (not a problem here, just for the general case).

    Of course, some of those problems can be avoided by appropriate use of std::weak_ptr.

Some more resources:

  • 1
    \$\begingroup\$ I would prefer turning a reference from left/right as it makes it clear that it is not a pointer that you should store because it can be destroyed at any point \$\endgroup\$
    – Emily L.
    Nov 19, 2017 at 10:53
  • 1
    \$\begingroup\$ @EmilyL.: The node might be nullptr, though, so you'd return a dangling reference in those cases. Sadly, there are no std::optional<T&> :/ \$\endgroup\$
    – hoffmale
    Nov 19, 2017 at 11:09
  • 1
    \$\begingroup\$ Now that you mention it yes. But that's also a problem with ops current code that directly dereferences and copy constructs from a possibly null-ptr. \$\endgroup\$
    – Emily L.
    Nov 19, 2017 at 20:07
  • \$\begingroup\$ @EmilyL.: Oh, yeah. I was so taken in with all the extra copies that I completely overlooked the nullptr dereferences. \$\endgroup\$
    – hoffmale
    Nov 20, 2017 at 7:30

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