Largest and Second Largest In BST

SecondLargest.java

private static int findLargest(BinaryTreeNode rootNode) {
    if (rootNode == null) {
        throw new IllegalArgumentException("Tree must have at least 1 node");
    }
    if (rootNode.right != null)  {
        return findLargest(rootNode.right);
    }
    return rootNode.value;
}

public static int findSecondLargest(BinaryTreeNode rootNode) {
    if (rootNode == null || (rootNode.left == null 
            && rootNode.right == null)) {
        throw new IllegalArgumentException("Tree must have at least 2 nodes");
    }

    // case: we're currently at largest, and largest has a left subtree,
    // so 2nd largest is largest in said subtree
    if (rootNode.left != null && rootNode.right == null) {
        return findLargest(rootNode.left);
    }

    // case: we're at parent of largest, and largest has no left subtree,
    // so 2nd largest must be current node
    if (rootNode.right != null
            && rootNode.right.left == null
            && rootNode.right.right == null) {
        return rootNode.value;
    }

    // otherwise: step right
    return findSecondLargest(rootNode.right);
}