Question: Is there any regular expression pattern to change this string

> 
> This is a mix string of üößñ and English. üößñ üößñ are Unicode words.
> 
> 
> 

to this?

> 
> This is a mix string of, üößñ, and English., üößñ üößñ, are Unicode words.
> 
> 
> 

Actually, I want to split English words and non-English words with comma.

Thanks.

Answer: 
It depends on whether you want a [**shallow** or **deep**](http://www.uow.edu.au/~lukes/TEXTBOOK/notes-cpp/oop-condestructors/shallowdeepcopy.html) copy. Assuming a deep copy, you need to be able to copy whatever's at the "leaves" hanging off a `TreeNode` object; so ideally the functionality should be in `TreeNode` (unless `Tree` is a friend class of `TreeNode` that you've designed to be deeply familiar with its implementation, which is often the case of course;-). Assuming something like...:

```
template <class Leaf>
class TreeNode {
  private:
    bool isLeaf;
    Leaf* leafValue;
    TreeNode *leftPtr, *rightPtr;
    TreeNode(const&Leaf leafValue);
    TreeNode(const TreeNode *left, const TreeNode *right);
  ...

```

then you could add to it a

```
  public:
    TreeNode<Leaf>* clone() const {
      if (isLeaf) return new TreeNode<Leaf>(*leafValue);
      return new TreeNode<Leaf>(
        leftPtr? leftPtr->clone() : NULL,
        rightPtr? rightPtr->clone() : NULL,
      );
    }

```

If `Tree` is taking care of this level of functionality (as a friend class), then obviously you'll have the exact equivalent but with the node being cloned as an explicit arg.
Two basic options:

If you have an iterator available, you can simply iterate over the elements in the tree and insert each one manually, as R. Pate described. If your tree class doesn't take explicit measures to balance the tree (e.g. AVL or red-black rotations), you'll end up effectively with a linked list of nodes this way (that is, all the left child pointers will be null). If you are balancing your tree, you'll effectively do the balancing work twice (since you already had to figure it out on the source tree from which you're copying).

A quicker but messier and more error-prone solution would be to build the copy top down by doing a breadth-first or depth-first traversal of the source tree structure. You wouldn't need any balancing rotations and you'd end up with an identical node topology.