Routes Between Node, Cracking the coding interview

Question

I am going though Cracking the coding interview and trying to improve my coding interview skills.

The question is "Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

I think the run time is N^2 but I could be wrong.

Is there a better way to do BFS?

package TreesAndGraph;

import java.util.LinkedList;
import java.util.Queue;

public class PathSearch
{
    public static void main(String[] args)
    {
        int[][] matrix = {
                {0, 1, 0, 0, 1, 1},
                {0, 0, 0, 1, 1, 0},
                {0, 1, 0, 0, 0, 0},
                {0, 0, 1, 0, 1, 0},
                {0, 0, 0, 0, 0, 0},
                {0, 0, 0, 0, 0, 0}};

        //Following are all true.
        System.out.println("0 -> 1: " + routeBetweenNode(matrix, 0, 1));
        System.out.println("0 -> 4: " + routeBetweenNode(matrix, 0, 4));
        System.out.println("0 -> 5: " + routeBetweenNode(matrix, 0, 5));
        System.out.println("0 -> 3: " + routeBetweenNode(matrix, 0, 3));
        System.out.println("0 -> 2: " + routeBetweenNode(matrix, 0, 2));
        System.out.println("1 -> 1: " + routeBetweenNode(matrix, 1, 1));
        System.out.println("2 -> 4: " + routeBetweenNode(matrix, 2, 4));
        System.out.println("1 -> 2: " + routeBetweenNode(matrix, 1, 2));

    }

    private static boolean routeBetweenNode(int[][] matrix, int start, int end)
    {
        Queue<Integer> q = new LinkedList<>();
        if (traverseMatrix(end, q, matrix[start])) return true;

        while (!q.isEmpty())
        {
            Integer activeVisited = q.remove();
            if (traverseMatrix(end, q, matrix[activeVisited])) return true;
        }
        return false;
    }

    private static boolean traverseMatrix(int end, Queue<Integer> q, int[] currentArray)
    {
        for (int i = 0; i < currentArray.length; i++)
        {
            int value = currentArray[i];
            if (value == 1)
            {
                q.add(i);
                if (i == end)
                {
                    return true;
                }
            }
        }
        return false;
    }
}