how count number of branches, in case branches integers. here's have far. seems work couple of cases.
public int evenbranches() { return evenbranches(overallroot); } private int evenbranches(inttreenode root) { if (root == null) { return 0; } int val = 0; if (root.left != null) { val += evenbranches(root.left); } else if (root.right != null) { val += evenbranches(root.right); } if (root.data % 2 == 0) { return val + 1; } else { return val; } } 
you can modify evenbranches() method below: think cover edge cases, if testcase left, let me know, fix it.
public int evenbranches() { return evenbranches(overallroot, 0); } private int evenbranches(inttreenode root, int count) { if(root == null || (root.left == null && root.right == null)) { return count; } if(root.data % 2 == 0) { count++; } count += evenbranches(root.left, count); count += evenbranches(root.right, count); return count; } 