count all nodes in a bst

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int countNodes(TreeNode root) {
        // Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
        // int count = 0;
        // TreeNode curr = root;
        // while(curr != null || !stack.isEmpty()){
        //     while(curr != null){
        //         stack.push(curr);
        //         curr = curr.left;
        //     }
        //     count++;
        //     curr = stack.pop();
        //     curr = curr.right;
        // }
        // return count;

        /**
        So the bellow implementation teaches you that when youre carrying through an addable
        number you dont need a new method to include it in the signature para pasarlo,
        you can just make it en la funcion y lo seteas a 1 y le sumas a esa variable el
        resultado de la proxima iteracion. base case es cuando sea nulo, return 0;
        this way you dont have to only do half the tree and ending it with a return.
        you can just add all of it to the 'global' variable and return that in the end
        from the first call;
        **/

        //current sum (assuming root is not null)

	int sum = 1;
	//base case
	if (root == null) return 0;
	//check left
	if (root.left != null) {
		//add left subtree to sum
		sum += countNodes(root.left);
	}
	//check right
	if (root.right != null) {
		//add right subtree to sum
		sum += countNodes(root.right);
	}
	//result
	return sum;
    }
}