Well, the whole point of the blog post is to argue for a new kind of syntactic sugar and the author explicitly mentions iterators.

> Well a range based for loop requires that your tree exist in memory AND that you have an iterator defined for your tree. With for_tree you could operate on an entirely imperative tree, without needing to define any iterators or generator functions. Here's an example where I'm checking every single string composed of "a", "b", and "c" of length 8 or less.

    for_tree(string x = ""; x.size() <= 8; x : {x+"a", x+"b", x+"c"}){
     print(x);
    }

You could definitely find every string composed of "a", "b", and "c" of length 8 or less by defining a custom iterator but it would be a verbose and unpleasant way of writing it:

    class StringIterator {
    public:
        using iterator_category = std::forward_iterator_tag;
        using value_type = std::string;
        using difference_type = std::ptrdiff_t;
        using pointer = const std::string*;
        using reference = const std::string&;

        StringIterator(bool begin = false) : is_end_(!begin) { if (begin) s_ = ""; }

        const std::string& operator*() const {
            if (is_end_) throw std::out_of_range("End iterator");
            return s_;
        }

        StringIterator& operator++() {
            if (is_end_) return *this;
            if (s_.size() < 8) return s_.push_back('a'), *this;
            while (!s_.empty() && s_.back() == 'c') s_.pop_back();
            if (s_.empty()) is_end_ = true;
            else s_.back() = s_.back() == 'a' ? 'b' : 'c';
            return *this;
        }

        StringIterator operator++(int) { auto tmp = *this; ++(*this); return tmp; }

        bool operator==(const StringIterator& other) const {
            return is_end_ == other.is_end_ && (is_end_ || s_ == other.s_);
        }

        bool operator!=(const StringIterator& other) const { return !(*this == other); }

    private:
        std::string s_;
        bool is_end_;
    };

    int main() {
        StringIterator begin(true), end;
        int count = 0;
        for (auto it = begin; it != end; ++it) ++count;
        std::cout << (count == 9841 ? "Pass" : "Fail") << std::endl;
        return 0;
    }