Function 检查二叉树是否平衡 C++

Question

I'm trying to implement a function that uses recursion to check if a binary search tree is balanced.

The template for the function I am using is

template<class T>
int BST<T>::is_balanced(BSTNode<T> *p) const
{

    if (p == 0) // Base case
        return 0;
    else {

}
}

Ive also created a function that checks the number of leaf nodes in the tree

template<class T>
int BST<T>::number_of_leaves(BSTNode<T>* start) const 
{
    if(start == NULL)
    {
        return 0;
    }
    if(start->right == NULL && start->left==NULL)
    {
        return 1;
    }
    return number_of_leaves(start->right) + number_of_leaves(start-> left);
}

From what I've read and saw, there needs to be something that gets the height of the left and right nodes. Is the second function able to be used for that purpose or am I overlooking something.

Any help to get this solved would be appreciated as all my attempts have not worked.

if needed this is the BSTNODE class

template<class T>
class BSTNode {
public:
    BSTNode() { left = right = 0; }
    BSTNode(const T& e, BSTNode<T> *l = 0, BSTNode<T> *r = 0) 
        { el = e, left = l, right = r; }
    T el;
    BSTNode<T> *left, *right;
};

Answer 1

Ended up working it out. I had confused myself by thinking I needed another function for the height. Below is the code i used

template<class T>
int BST<T>::is_balanced(BSTNode<T>* root) const
{
        if(root == NULL)
        return 0;
        else {
                int lh = is_balanced(root->left);
                int rh= is_balanced(root->right);
                if (lh == -1 || rh == -1) return -1;
        if (abs(lh-rh) > 1) return -1;
        if(lh > rh) return lh +1;
        return rh+1;
        }
}