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

The above wont work, consider a tree that has a root value for 5, left is 4, right is 6, left of the left is 3 and right of the left is 6. This one should work (for integers): boolean isBST(Node root) { return isBST_helper(root, INT_MIN, INT_MAX); } boolean isBST_helper(Node root, int min, int max) { if (root == null) return true; return root.value >= min && root.value < max && isBST(root.left, min, root.value) && isBST(root.right, root.value+1, max); }

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 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)