Knowing that the inorder traversal of BSTs produces sorted lists, write an iterative function function that given 2 BSTs, the function checks whether the elements of both BSTs are the same or not. The time complexity of your algorithm should be \(O(max(m, n))\), where \(m\) and \(n\) are the numbers of elements in the two BSTs . 

int samei(Btree A, Btree B)
{
	stack s1 = CreateStack();
	stack s2 = CreateStack();
	int done1 = 0;
	int done2 = 0;
	int v1, v2;


	while (!done1 && !done2){
		while (!done1){
			if (A){
				Push(&s1, A);
				A = A->left;
			}
			else{
				if (Top(s1, &A)){
					Pop(&s1);
					v1 = A->data;
					A = A->right;
					break;
				}
				else
					done1 = 1;
			}
		}
		while (!done2){
			if (B){
				Push(&s2, B);
				B = B->left;
			}
			else{
				if (Top(s2, &B)){
					Pop(&s2);
					v2 = B->data;
					B = B->right;
					break;
				}
				else
					done2 = 1;
			}
		}
		if (done1 == 1 && done2 == 1)	return 1;
		if (v1 != v2) return 0;
		done1 = 0;
		done2 = 0;
	}
}