Write a recursive method isBinaryTree() that takes a Node as an argument and returns true if the subtree count field N is consistent in the data structure rooted at that node, false otherwise.

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public class Node , Value> Key key; // key Value val; // associated value NodeKey, Value> left, right; // links to subtrees static int lastvalue = Integer.MIN.VALUE; in n; // # nodes in subtree rooted here public Node (Key key, Value val, int n) this.keykey; this.val-val; k 3 @param node root of the sub-tree @return true if for each node in the sub-tree, n field value matches the number of nodes in its sub-tree, otherwise false public static boolean isBinaryTree(Node node) { return isBST (node, null, null); public static , Value boolean isBST (Node node, Key low, Key high) if (nodenul1) ( return true; if (lownull && low.compareTo(node.key) ) return false if (high != null && high . compareTo(node . key) 0) { return Talse return isBST (node.left, low, node.key) && isBST(node.right, node.key, high); public class Node , Value> Key key; // key Value val; // associated value NodeKey, Value> left, right; // links to subtrees static int lastvalue = Integer.MIN.VALUE; in n; // # nodes in subtree rooted here public Node (Key key, Value val, int n) this.keykey; this.val-val; k 3 @param node root of the sub-tree @return true if for each node in the sub-tree, n field value matches the number of nodes in its sub-tree, otherwise false public static boolean isBinaryTree(Node node) { return isBST (node, null, null); public static , Value boolean isBST (Node node, Key low, Key high) if (nodenul1) ( return true; if (lownull && low.compareTo(node.key) ) return false if (high != null && high . compareTo(node . key) 0) { return Talse return isBST (node.left, low, node.key) && isBST(node.right, node.key, high)