Diameter of Binary Tree - Better Design

Question

I have written a code for finding diameter of Binary Tree. Need suggestions for the following:

1. Can I do this without using static variable at class level?

2. Is the algorithm fine/any suggestions?

   public class DiameterOfTree {   
   public static int diameter = 0; 
   public static int getDiameter(BinaryTreeNode root) {        
       if (root != null) {                     
           int leftCount = getDiameter(root.getLeft());
           int rightCount = getDiameter(root.getRight());
           if (leftCount + rightCount > diameter) {
               diameter = leftCount + rightCount;
               System.out.println("---diameter------------->" + diameter);
           }           
           if ( leftCount > rightCount) {
               return leftCount + 1;
           }
           return rightCount + 1;
       }
       return 0;
     }
   }
   

Answer 1

There are three cases to consider when trying to find the longest path between two nodes in a binary tree (diameter):

1. The longest path passes through the root,
2. The longest path is entirely contained in the left sub-tree,
3. The longest path is entirely contained in the right sub-tree.

The longest path through the root is simply the sum of the heights of the left and right sub-trees (+1 for the root not necessary since the diameter of a tree with a root node and 1 left, 1 right subtree nodes will be 2), and the other two can be found recursively:

public static int getDiameter(BinaryTreeNode root) {        
    if (root == null)
        return 0;

    int rootDiameter = getHeight(root.getLeft()) + getHeight(root.getRight()); //Removing the +1
    int leftDiameter = getDiameter(root.getLeft());
    int rightDiameter = getDiameter(root.getRight());

    return Math.max(rootDiameter, Math.max(leftDiameter, rightDiameter));
}

public static int getHeight(BinaryTreeNode root) {
    if (root == null)
        return 0;

    return Math.max(getHeight(root.getLeft()), getHeight(root.getRight())) + 1;
}

Answer 2

Here is an O(n) solution with minimal changes to the accepted answer:

public static int[] getDiameter(BinaryTreeNode root) {
    int[] result = new int[]{0,0};    //1st element: diameter, 2nd: height    
    if (root == null)  return result;
    int[] leftResult = getDiameter(root.getLeft());
    int[] rightResult = getDiameter(root.getRight());
    int height = Math.max(leftResult[1], rightResult[1]) + 1;
    int rootDiameter = leftResult[1] + rightResult[1] + 1;
    int leftDiameter = leftResult[0];
    int rightDiameter = rightResult[0];
    result[0] = Math.max(rootDiameter, Math.max(leftDiameter, rightDiameter));
    result[1] = height;

    return result;
}

It just calculates height and diameter at the same time. And since Java does not have pass-by-reference I defined an int[] to return the result.