Diameter of Binary Tree - Better Design

You don't need to store the result in the static field diameter. Simply use the static method like that:

public class DiameterOfTree {

    public static long getDiameter(BinaryTreeNode root) {
        if (root != null) {
            long leftDiameter = getDiameter(root.getLeft());
            long rightDiameter = getDiameter(root.getRight());
            long leftHeight = getHeight(root.getLeft());
            long rightHeight = getHeight(root.getRight());
            return Math.max(leftHeight + rightHeight + 1, Math.max(leftDiameter, rightDiameter));
        }
        return 0;
    }

    public static long getHeight(BinaryTreeNode root) {
        if (root != null) {
            long leftHeight = getHeight(root.getLeft());
            long rightHeight = getHeight(root.getRight());
            return  1 + Math.max(leftHeight, rightHeight);
        }
        return 0;
    }
}

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;
}

Here is a solution in Java that has O(N) time complexity. It calculates the height in the same recursion when calculating the diameter. Reference Link

private class HeightWrapper {
    int height = 0;
}

private int getDiameter_helper(BinaryTreeNode root, HeightWrapper wrapper) {
    if (root == null) {
        return 0; // diameter and height are 0
    }

    /* wrappers for heights of the left and right subtrees */
    HeightWrapper lhWrapper = new HeightWrapper();
    HeightWrapper rhWrapper = new HeightWrapper();

    /* get heights of left and right subtrees and their diameters */
    int leftDiameter = getDiameter_helper(root.left, lhWrapper);
    int rightDiameter = getDiameter_helper(root.right, rhWrapper);

    /* calculate root diameter */
    int rootDiameter = lhWrapper.height + rhWrapper.height + 1;

    /* calculate height of current node */
    wrapper.height = Math.max(lhWrapper.height, rhWrapper.height) + 1;

    /* calculate the diameter */
    return Math.max(rootDiameter, Math.max(leftDiameter, rightDiameter));
}

public int getDiameter(BinaryTreeNode root) {
    HeightWrapper wrapper = new HeightWrapper();
    return getDiameter_helper(root, wrapper);
}

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.