从二叉搜索树C++返回引用

考虑我的二叉搜索树中的以下搜索函数。

template <class elemType>
elemType& BSTree<elemType>::search(const elemType & searchItem) const
{
    std::cout << "in 1st teir search" << std::endl;
    if (root == NULL)
    {
        std::cout << "Tree is empty, and there for no data will be in this tree." << std::endl;
    }
    else
    {
        std::cout << "Entering 2nd teir search" << std::endl;
        return search(root, searchItem);
    } //End else
} //End search(1param)
template <class elemType>
elemType& BSTree<elemType>::search(nodeType<elemType>* node, const elemType& dataToFind) const
{
    elemType found;
    if (node == NULL)
    {
        std::cout << "Not found. Node is null." << std::endl;
    }
    else
    {
        if (node->data == dataToFind)
        {
            std::cout << "Data found" << std::endl;
            found = node->data;
        }
        else if (node->data < dataToFind)
        {
            std::cout << "Data not found, searching to the RIGHT" << std::endl;
            found = search(node->rLink, dataToFind);
        }
        else
        {
            std::cout << "Data not found, searching to the LEFT" << std::endl;
            found = search(node->lLink, dataToFind);
        }
    } //End else
    return found;
} //End search(2param)

每当我访问/搜索不是root的数据时，当我去分配该数据时，我的程序就会崩溃。

我错过了什么？

注意：理解也许我可以在遍历中使用函数指针来返回值，但出于目的，我使用我的树进行搜索将返回对对象的引用。