Why is the root always null in this method insert method for binary tree

Question

Iam trying to implement a recursive insert node method for BST class

 public void insertNode(Node r, Node n)
{
    if(r == null)
    {
        System.out.println("r=n"+ n.data);
        r = n;
    }
    else
    {
        System.out.println("r=! null finding place in tree for "+ n.data);
        if(n.data <= r.data)
        {
            if(r.left == null)
                r.left = n;
            else
                insertNode(r.left, n);
        }
        else
        {
            if(r.right == null)
                r.right = n;
            else
                insertNode(r.right, n);
        }
    }
}

Im trying to invoke this method like so:

  int[] arrTree = {34, 2, 56, 12, 44, 39, 56, 1};
    BT t = new BT();
    for (int i = 0; i < arrTree.length; i++)
    {
        //System.out.println("Tree Root = "+ t.getRoot());
        BT.Node n =  t.new Node(arrTree[i]);
        t.insertNode(t.root, n);
    }

But Im always getting this output:

 r=n34
 r=n2
 r=n56
 r=n12
 r=n44
 r=n39
 r=n56
 r=n1

Node is an inner class of BT.

I cannot figure it out after hours of running and trying different things what Im doing wrong.

Answer 1

Judging by the code you've shown, my money's on the error being here:

public void insertNode(Node r, Node n)
{
    if(r == null)
    {
        System.out.println("r=n"+ n.data);
        r = n; //you overwrite the value of r but never use it
    }

Node r is actually a separate reference to whatever t.root refers to, so replacing r with another value won't change whatever t.root refers to outside the method. You can modify the referenced data inside a method, but not the reference itself.

Answer 2

Java is pass-by value language. In case of objects you pass the value of the reference, thus, creating new reference.

public void insertNode(Node r, Node n) {
    if(r == null) {
        r = n;
    }

Here you substitute the new reference to the root with the other reference; original reference to root (t.root) left unchanged.

To fix this issue you can get rid of the first parameter in the insertNode() method - the tree's root is the part of its implementation, so the tree already knows the reference to its root. Change all rs inside insertNode() to this.root.

Answer 3

When you set r within the method, it does not affect the node that you pass in. So t.root is never set.