如何实现返回二叉树中存储着偶数的分支节点数

Java中如何实现返回二叉树中存储着偶数的分支节点数？图中的代码又应该如何补全？

//假设IntTreeNode的结构如下
class IntTreeNode {
    int val;//值
    IntTreeNode left;//左子树
    IntTreeNode right;//右子树

    public IntTreeNode(int val) {
        this.val = val;
    }
}

//补全后的代码
private int countEvenBranches(IntTreeNode root) {
        if (root == null || root.left == null && root.right == null && root.val % 2 != 0) {
            return 0;
        } else {
            int result = 0;
            if (root.val % 2 == 0) {
                result = 1;
            }
            return result + countEvenBranches(root.left) + countEvenBranches(root.right);
        }
    }

public class Web_02_countEvenBranches {

    public static void main(String[] args) {
        TreeNode tree = TreeNode.createTree();
        TreeNode.printTree(tree);
        int i = countEvenBranches(tree);
        System.out.println(i);
    }

    //递归求解
    private static int countEvenBranches(TreeNode root) {
        //不满足条件直接返回
        if (root == null || root.left == null && root.right == null && root.val % 2 != 0) {
            return 0;
        } else {
            //返回当前节点的result
            int result = 0;
            //满足
            if (root.val % 2 == 0) {
                result = 1;
            }
            //根节点+左右节点
            return result + countEvenBranches(root.left) + countEvenBranches(root.right);
        }
    }

}