Java : Count even values in a Binary Search Tree recursively

Question

Hi,

I need to find out how many even values are contained in a binary tree.

this is my code.

private int countEven(BSTNode root){

if ((root == null)|| (root.value%2==1))
return 0;

return 1+ countEven(root.left) + countEven(root.right);


}

this i just coded as i do not have a way to test this out. I'm not able to test it out at the moment but need an answer so badly. any help is deeply appreciated.

Answer 1

private int countEven(BSTNode root) {
   if (root == null)
      return 0;

   int n = countEven(root.left) + countEven(root.right);
   if(root.value % 2 == 0)
      return n + 1;
   else
      return n;
}

Answer 2

If there is an node with an odd value containing subnodes with even values, the subnodes will not be counted in your code. Small enhancement below.

private int countEven(BSTNode root){

  if (root == null)
    return 0;

  int val = (root.value%2==1) ? 0 : 1;

  return val + countEven(root.left) + countEven(root.right);


}