package test; public class CheckBST { static boolean isBST = false; static boolean continueCheck = true; private static void checkBST(Node root) { if (continueCheck) { Node left = root.getLeftNode(); Node right = root.getRightNode(); if (left.getValue() < root.getValue() && root.getValue() < right.getValue()) { isBST = true; if (null != left.getLeftNode() || null != left.getRightNode()) { checkBST(left); } if (null != right.getLeftNode() || null != right.getRightNode()) { checkBST(right); } } else { isBST = false; continueCheck = false; } } } public static void main(String[] args) { Node left = new Node(2, new Node(1), new Node(3)); Node right = new Node(6, new Node(9), new Node(7)); Node root = new Node(4, left, right, true); checkBST(root); System.out.println("is BST ? ::" + isBST); } }

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How would you validate if its not a BST because it has more than 2 children, using the recursion algo above?