Count Good Nodes in Binary Tree

lcpy gen -n 1448   # by problem number
lcpy gen -s count_good_nodes_in_binary_tree   # by problem name

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.

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

Input: root = [1]
Output: 1
Explanation: Root is considered as good.

from leetcode_py import TreeNode


class Solution:
    # Time: O(n)
    # Space: O(h) - recursion stack, h = tree height
    def good_nodes(self, root: TreeNode[int] | None) -> int:
        if root is None:
            return 0

        def dfs(node: TreeNode[int], max_so_far: int) -> int:
            good = 1 if node.val >= max_so_far else 0
            next_max = max(max_so_far, node.val)
            total = good
            if node.left is not None:
                total += dfs(node.left, next_max)
            if node.right is not None:
                total += dfs(node.right, next_max)
            return total

        return dfs(root, root.val)