Java Programming

I'm getting "It is a bad idea to assign a new value to a formal parameter within a method. Consider using a local variable instead." How can I set this up properly also?

Here's how to delete a node from a BST:
[http://en.wikipedia.org/wiki/Binary_search_tree#Deletion]

Updated code, however it doesn't work too well. Any suggestions?
That wiki didn't help too much.

public void delete( int myVal )
    {
        delete( root, myVal );
    }

private void delete( Tnode node, int value )
    {
        if ( ( node.lkid == null ) && ( node.rkid == null ) )
        {
            node = null;
        }
        else if ( ( node.lkid == null ) && ( node.rkid != null ) )
        {
            node = node.rkid;
        }
        else if ( ( node.lkid != null ) && ( node.rkid == null ) )
        {
            node = node.lkid;
        }
        else if ( ( node.lkid != null ) && ( node.rkid != null ) )
        {
            if ( node.lkid.val > node.rkid.val )
            {
                node = node.lkid;
                node.lkid = null;
            }
            else if ( node.rkid.val > node.lkid.val )
            {
                node = node.rkid;
                node.rkid = null;
            }
        }
    }

Sorry to hear that, but I can't make it clearer than on that wiki.

Nevermind, figured it out.

Good to hear it.
Would you mind posting your solution here as well then? I mean if someone else is trying to find the answer to your question and, finds this post after searching the web, s/he doesn't know the answer.

Sure, SOLUTION

public void delete( int myVal )
    {
        delete( root, myVal );
    }


    /**
     * Recursive delete method.
     * 
     * @param node
     *            the node given for recursion
     * @param value
     *            the value given for recursion
     */
    private void delete( Tnode node, int value )
    {
        if ( ( node.lkid == null ) && ( node.rkid == null ) )
        {
            this.root = null;
        }
        else if ( ( node.lkid == null ) && ( node.rkid != null ) )
        {
            this.root = node.rkid;
        }
        else if ( ( node.lkid != null ) && ( node.rkid == null ) )
        {
            this.root = node.lkid;
        }
        else if ( ( node.lkid != null ) && ( node.rkid != null ) )
        {
            if ( node.lkid.val > node.rkid.val )
            {
                this.root = node.lkid;
                node.lkid = null;
            }
            else if ( node.rkid.val > node.lkid.val )
            {
                this.root = node.rkid;
                node.rkid = null;
            }
        }
    }

That's no valid delete(...) method!
You're not making a recursive call anywhere, so you won't be able to get to nodes that are more than one "hop" from the root. And even if you don't need a recursive call, like with the following tree:

-
     3
    / \
   /   \
  2     5
 / \   / \
*   * *   *

// * == null

and you'd call delete(2), then your method would leave the tree like this:

-
  5
 / \
*   *

which can't be correct, right?

Yes, I just realized after testing that that did not work.

Any suggestions on fixing this?

Like I said: if you can't follow the explanation on the wiki page, I don't think I can help you further. But, this is how a recursive delete method could look like:

public void remove('valueToDelete') {
  // assume the tree is not empty and 'valueToDelete' is present in this tree
  this.remove('valueToDelete', 'root', null);
}
    
private void remove('valueToDelete', 'current', 'parent') {

  'diff' <- the difference between 'valueToDelete' and 'current.value'

  if('diff' is more than 0) {
    // recursively traverse into the right sub-tree of 'current'
  } 
  else if('difff' is less than 0) {
    // recursively traverse into the left sub-tree of 'current'
  } 
  else { // Ok, we found the node.

    'tempNode' <- the replacement of 'current'
    // Special cases: see the wiki page (here you need te assign 'tempNode')  
    if('current' has only one child) {
      // your code
    } 
    else if('current' has two children) {
      // your code
    }
    else { // no children
      // your code
    }
    
    // Rebuild the new tree (replace 'current' with 'tempNode')   
    if('current' has no parent) {
      // your code
    } 
    else if('current' is the left child of 'parent') {
      // your code
    } 
    else { // 'current' is the right child of 'parent'
      // your code
    }
  }
}