"""
1448. Count Good Nodes in Binary Tree
Difficulty: Medium
https://leetcode.com/problems/count-good-nodes-in-binary-tree/

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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].
"""


# 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:
        def dfs(node, max_val):
            if not node:
                return 0

            is_good = 1 if node.val >= max_val else 0
            new_max = max(max_val, node.val)

            return is_good + dfs(node.left, new_max) + dfs(node.right, new_max)

        return dfs(root, root.val)