I found an interesting problem on https://leetcode.com/problems/count-good-nodes-in-binary-tree/:
Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
Return the number of good nodes in the binary tree.
 
Example 1:

Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.
Example 2:

Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.
Example 3:
Input: root = [1]
Output: 1
Explanation: Root is considered as good.
 
Constraints:

The number of nodes in the binary tree is in the range [1, 10^5].
Each node's value is between [-10^4, 10^4].

I tried it in Python, but could not do it. Can you solve it?
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def goodNodes(self, root: TreeNode) -> int:
        return self.count_good_nodes(root, root.val)
    
    def count_good_nodes(self, root: TreeNode, highest_val: int) -> int:
        current_highest = highest_val
        total = 0
        
        if root.val >= current_highest:
            total += 1
            current_highest = root.val

        if root.left:
            total += self.count_good_nodes(root.left, current_highest)
        if root.right:
            total += self.count_good_nodes(root.right, current_highest)
        
        return total