find the node in k distance in a given tree list, solution 2 : reverse back complexity = o (n)

gistfile1.java

public class HelloWorld {

    public static void main(String[] args) {
        Node r = new Node(0);

        r.left = new Node(1);
        r.right = new Node(2);

        r.left.left = new Node(3);
        r.left.right = new Node(4);
        r.left.left.left = new Node(5);

        r.right.right = new Node(6);
        r.right.left = new Node(7);
        r.right.left.right = new Node(8);


        find(r, 2);
    }

    public static class Node {
        Node parent;
        Node left;
        Node right;
        int value;

        public Node(int value) {
            this.value = value;
        }

    }


    public static void find(Node n, int k) {
        
        if (n.left == null && n.right == null) {
            
            int x = 0;
            Node t = n;
            while (x < k){
                t = t.parent;
                x++;
                if (t != null && x == k){
                     System.out.println("leaf : " + n.value + " node in " + k + " distance : " + t.value);
                }
            }
        }

        if (n.left != null) {
            n.left.parent = n;
            find(n.left, k);
        }

        if (n.right != null) {
            n.right.parent = n;
            find(n.right, k);
        }



    }

}