Topological sort in Java
class Graph {
    public int vertices;
    LinkedList&lt;Node&gt; nodes = new LinkedList&lt;Node&gt;();
}


Graph

Graph, as other data structures in general, should be declared 
public. Because you want them to be able to be used outside of the 
package they are declared in. 

nodes, vertices should be private so that you can know that they 
are not changed outside the Graph class. 

nodes should not be a LinkedList. You must depend on abstractions as 
much as possible. You should prefer interfaces such as List over 
specific implementations LinkedList.

Nodes of a graph is not a List, it is a Set. A standard graph cannot 
have multiple copies of a node. You should prefer a Set to represent a 
set unless you have a good reason. 

Node

All of the above points also apply to Node. Apart from those: 

AdjacenctNode should be named adjacentNode by Java naming 
convention. 

Feel free to remove parameterless call to super();, although Eclipse 
adds it by default, it's just noise. 

TopologicalSortGraph

Remove unused code : 

TopologicalSortGraph Graph = new TopologicalSortGraph();


Always remove commented code. If you need to see previous versions of a 
code use a version control system. : 

//public LinkedList&lt;Node&gt; nodes  = new LinkedList&lt;Node&gt;();


Do not put more than one space between tokens, use autoformat of your 
IDE to fix formatting after changing a piece of code: 

public static void  topologicalSort(Graph graph) {


The snippets :

if(child.inDegree==0){
    q.add(child);
    ++vertexProcessesCtr;
    System.out.println(child.data);
}


and     

if(m.inDegree==0){
    ++vertexProcessesCtr;
    q.add(m);
    System.out.println(m.data);
}


are duplicates. They should be extracted to a private method. You are 
missing some kind of abstraction there. 

You are changing the internals of an object passed in as a parameter: 
--child.inDegree; I do not expect my graph to change after I ask to 
see its nodes printed in topological order. What if I want to print 
them again?

You are mixing the calculation and the printing out of the calculation result, 
I do not expect to see printlns in a method implementing an algorithm: 

System.out.println( .... );

What if I want the result to be printed somewhere other than System.out? 
What if I do not want the result to be printed at all and want it to be 
used as an intermediate step in a bigger calculation instead? 

You probably want to return a List&lt;Node&gt; from a topological sort 
algorithm. List is the standard return type when you are sorting some 
collection, that is when the result is a collection whose order is 
important. You can then print that list as many times as you want or 
pass it as a parameter to wherever you like. 

public static void main(String[] args) {

You should separate test code from your main code. If your actual class 
is named TopologicalSortGraph put your test code in 
TopologicalSortGraphTest. 

Use de facto standard JUnit so that instead of one big main you can 
have many small tests. You can run any one of them or all of them easily 
from within your IDE or from the command line. 

You should try to separate the test code into separate source 
directories or even into separate projects. Your implementation code 
should not need or know about your test code to compile. 

Another spacing (indentation) problem : 

Node TEN = new Node("10");


Your code should align well. So that it reads neatly top to bottom and 
scopes in it are easily identified. 

TEN should be ten by Java naming convention.

Instead of these two you should use your addAdjNode method instead:

SEVEN.AdjacenctNode.add(ELEVEN);
ELEVEN.inDegree++;


Also access chains like node.AdjacenctNode.add(otherNode) or x.y.z 
are usually a sign that your encapsulation is not good enough. In this 
case you are modifying a collection of one class from another class. 
It's a problem waiting to happen. Same encapsulation problem is present 
in ELEVEN.inDegree++, too. 

The root problem here is adjacency is a property of the graph -Remember 
G = (V, E) from school?- and not of the nodes themselves. Instead of 
node.addAdjNode(otherNode), you should use a method like 
graph.addEdge(node, otherNode). 

Same problem also exists here: g.nodes.add(TWO); You should have a
Graph.addNode(node) method instead. 

Coming back to G = (V, E); it says, if you listen carefully, you need 
a set of vertices and a set of edges to have a graph and should not add 
nodes one by one. Ideally, Graph would have a constructor like this 
public Graph(Set&lt;Node&gt; vertices, Set&lt;Edge&gt; edges).