Decide If A Binary Tree Is A Binary Search Tree

Java:

Recursive solution 1:

class Node {
    public Node left, right;
    public int value;
    private boolean first;
    private int previous;
    public boolean isBST() {
        first = true;
        return isBST(this);
    }
    private boolean isBST(Node n) {
        if (n == null) return true;
        if (!isBST(n.left)) return false;
        if (first) {
            first = false;
            previous = n.value;
        } else if (previous >= n.value) return false;
        previous = n.value;
        return isBST(n.right);
    }
}

Recursive solution 2:

public boolean isBST() {
    return isBST(this, null, null);
}
public boolean isBST(Node n, Integer min, Integer max) {
    if (n == null) return true;
    if ((min != null && n.value <= min) || (max != null && n.value >= max)) return false;
    return isBST(n.left, min, n.value) && isBST(n.right, n.value, max);
}

Iterative solution:

public boolean isBST() {
    boolean first = false;
    int previous = 0;
    Stack<Node> nodes = new Stack<Node>();
    Node n = this;
    while (n != null) {
        nodes.push(n);
        n = n.left;
    }
    while (!nodes.empty()) {
        n = nodes.pop();
        if (first) first = false;
        else if (previous >= n.value) return false;
        previous = n.value;
        n = n.right;
        while (n != null) {
            nodes.push(n);
            n = n.left;
        }
    }
    return true;
}