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Binary Search Tree in Java

Hi I have a binary search tree that I'm building.  I have an array of coordinates (Point objects) that I need to insert.  
The tree should branch based on the X-coordinate on even levels
The tree should branch based on the Y-coordinate on odd levels

I have an array that stores the coordinates, I have two methods to sort the array: xSort and ySort

I'm having a hard time trying to figure out how to build the tree.  Here is what I have so far, but it's probably not the way I should do it.
public Node buildTree(Point[] _arr, int left, int right){
        	if(_arr.length==0)
        		return null;
        	if(_arr.length==1)
        		root = new Node(_arr[0]);
        		return root;
        	else{
        		int median = (left + right)/2;
        		xSort(_arr);
        		root = new Node(_arr[median]);
        		root.left = buildTree(_arr, left, median+1);
        		root.right = buildTree(_arr, median+1, right);
        		return root;
        	}
        }

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ubuntuguy
Asked:
ubuntuguy
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2 Solutions
 
PhasmidCommented:
I think what you really need is an R-tree (since you are working in 2-dimensions -- B-trees are essentially one dimensional).  There are lots of resources on the web (some with Java code).  I suggest you try here: http://gis.umb.no/gis/applets/rtree2/jdk1.1/
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ubuntuguyAuthor Commented:
Is this a good approach?  I get the following error:
Exception in thread "main" java.lang.Error: Unresolved compilation problem:
      Unreachable code
int median;
    	static Point [] leftPartition;
		static Point [] rightPartition;
        private static int calculateMedian(Point [] _arr){
        	if(_arr == null)
        		return -1;
        	if(_arr.length==0)
        		return 0;
        	else
        		return _arr.length/2;
        }
        private static Point[] calculateLeftPartition(Point [] _arr){
        	if(_arr == null)
        		return null;
        	else{
        		for(int i=0; i<calculateMedian(_arr); i++){
        			leftPartition[i] = _arr[i];
        		}
        	}
        	return leftPartition;
        }
        private static Point[] calculateRightPartition(Point [] _arr){
        	if(_arr == null)
        		return null;
        	else{
        		for(int i=calculateMedian(_arr)+1; i<_arr.length; i++){
        			rightPartition[i] = _arr[i];
        		}
        	}
        	return rightPartition;
        }
        
        
        public Node buildTree(Point[] _arr, Node _root){
        	if(_arr.length==0){
        		_root = new Node(null);
        		return _root;}
        	if(_arr.length==1){
        		_root = new Node(_arr[0]);
        		return _root;}
        	else{
        		xSort(_arr);
        		_root = new Node(_arr[calculateMedian(_arr)]);
        		ySort(calculateRightPartition(_arr));
        		ySort(calculateLeftPartition(_arr));
        		return buildTree(calculateLeftPartition(_arr), _root.left);
        		return buildTree(calculateRightPartition(_arr), _root.right);
        		}
        }
        
 
		public void xSort(Point [] arr){
			Arrays.sort(arr, new Point.CompX());
		}
		public void ySort(Point [] arr){
			Arrays.sort(arr, new Point.CompY());
		}

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PhasmidCommented:
Where did this bit of code come from?  Keep looking for Rtree (or R-tree) Java stuff on web until you find some code that works.
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Kevin CrossChief Technology OfficerCommented:
One thing is that you cannot instantiate Node.  If it is org.w3c.dom.Node or javax.xml.soap.Node, they are both Interfaces.  
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Kevin CrossChief Technology OfficerCommented:
But, you may have another Node implementation that I don't have on my system, so to answer your question this is the code that is unreachable.

return buildTree(calculateLeftPartition(_arr), _root.left);
            return buildTree(calculateRightPartition(_arr), _root.right);

Once you have returned from the method, the second return cannot be reached.
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ubuntuguyAuthor Commented:
Yes I'm using another Node implementation.  Why is
return buildTree(calculateLeftPartition(_arr), _root.left);
            return buildTree(calculateRightPartition(_arr), _root.right);
not reachable?
public class Node {
	
	public Point point;
	public Node left, right, parent;
	public int level;
	
	public Node(){  
		point = null;
		left = null;
		right = null;
                parent = null;
		level = 0;
	}
 
	public Node(Point _data){
		point = _data;
		point.x = _data.getX();
		point.y = _data.getY();
		left = null;
		right = null;
	}
	public Node(Point _data, Node _left, Node _right){
		point = _data;
		point.x = _data.getX();
		point.y = _data.getY();
		left = _left;
		right = _right;
	}
 
	public boolean isNull(Node _node){
		return _node.point==null;
	}
	
	public void setPoint(Point _point){
		this.point = _point;
	}
	public void setPoint(int a, int b){
		this.point.setX(a);
		this.point.setY(b);
		
	}
	public void setLeft(Node _node){
		this.left = _node;
	}
	public void setRight(Node _node){
		this.right = _node;
	}
	public void setLevel(int _level){
		this.level = _level;
	}
	
	public String toString(){
		if(this.point==null)
			return "Null Value";
		else 
			return this.point.toString();
	}
        }

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PhasmidCommented:
You can only make one return from a method.  If you want to return two separate objects, as it appears you might do here, then you would have to encapsulate them into a single object - or alternatively (and less elegantly) return them via method parameters.
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Kevin CrossChief Technology OfficerCommented:
Thought I made that clear above, you can return from different control structures like you have under if...then...else.  But in each section (path of execution), once you have returned to the caller you no longer have thread of execution to do any more code much less return anything else.

You can alter your buildTree method to return a List<Node> object instead of a Node.  For most of the returns this will just be a single element or null, but you can also have it return your mutliple Nodes like this:

List<Node> list = new ArrayList<Node>();
list.add(node1);
list.add(node2);
return list;

You can use an array two, just think List<T> is a little more elogant. :)  If there are limitations on what API you can use for class then you can use a simple array of Node[].
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Kevin CrossChief Technology OfficerCommented:
Looking at your code, you could just use this constructor:

public Node(Point _data, Node _left, Node _right){
            point = _data;
            point.x = _data.getX();
            point.y = _data.getY();
            left = _left;
            right = _right;
      }

It lets you define one Node object that contains the left Node and the right Node.  Then just return that Node.
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ubuntuguyAuthor Commented:
Sorry guys, I've taken a different approach.  This is how I'm doing it now:  It works for just branching on the X-value.  I'm unsure how to branch by X on even levels and by Y on odd levels.  I have already define my methods that sort the array by X or by X.  
	public static Node buildTree(Point[] _arr, int _left, int _right){
		Node _t;
		int median;
		if(_right<_left)
			return null;
		if(_right==_left){
			_t = new Node(_arr[_left], null, null);
			return _t;
		}
		median = (_left+_right)/2;
		_t = new Node(_arr[median], null, null);
		_t.left = buildTree(_arr, _left, median-1);
		_t.right = buildTree(_arr, median+1, _right);
		_t.size = _right-_left+1;
		return _t;
	}

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Kevin CrossChief Technology OfficerCommented:
Glad you found an approach that works.  But you did correct what we said which was the double return, so code should at least compile for you now.

Good luck.

Happy coding!

/kev
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