public boolean checkBst(Node n) { if (n == null) return true; if (checkBst(n.left)) { if ( checkBst(n.right)) { boolean isBst = false; if (n.left != null){ isBst = n.left.data n.data; } return isBst; } } return false; }

A mistake below corrected.. public boolean checkBst(Node n) { if (n == null) return true; if (checkBst(n.left)) { if ( checkBst(n.right)) { boolean isBst = true; if (n.left != null){ isBst = n.left.data n.data; } return isBst; } } return false; }

What Jordan is saying is correct. We need to maintain bound for left and right subtrees. The bound for left is min and root.value and for right is root.value and max. The above give n by me in incorrect. The correct on lines of Jordan - public static boolean isBST(Node t, int min, int max) { if (t == null) { return true; } if (isBST(t.left, min, t.value) && isBST(t.right, t.value, max)) { boolean bst = true; if (t.left != null) { bst = (t.left.value >= min) && (t.left.value = min) && (t.right.value <= max); } return bst; } else{ return false; } }

isBST(Node root) if root == null return false return isBST(root, root.value, true) && isBST(root, root.value, false) isBST(Node root, Int root_value, Boolean left) if root == null return true if left == false ? rot.value = root_value && (root.left == null || root.left.value <= root.value) && (root.right == null || root.value < root.right.value) return true && isBST(root.left, root_value, left) && isBST(root.right, root_value, left) return false

It is as if the simple solution above didn't detect correctly following case: 2 1 3 0. 4

isBST(Node node): if node == null: // no node is technically a bst tree return true; if node.left != null: if node.value node.right: return false; return isBST(node.left) && isBST(node.right); test: isBST(tree.root)