# 1448. Count Good Nodes in Binary Tree

#### Medium

***

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:**

![](https://assets.leetcode.com/uploads/2020/04/02/test_sample_1.png)

<pre><code>Input: root = [3,1,4,3,null,1,5]
<strong>Output:
</strong> 4
<strong>Explanation:
</strong> 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.
</code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/04/02/test_sample_2.png)

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

**Example 3:**

<pre><code>Input: root = [1]
<strong>Output:
</strong> 1
<strong>Explanation:
</strong> Root is considered as good.
</code></pre>

&#x20;

**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]`.

![](https://1865766684-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FLgwjo4Xadqnv6PXZEA0Z%2Fuploads%2FwaZknUXCmVMVE0Q8O9jE%2Fimage.png?alt=media\&token=39856dac-ebbb-48a1-b83e-1fab24dc5845)

```python
# 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.dfs(root, root.val)
    
    def dfs(self, root, val):
        if root is None:
            return 0
        result = 1 if root.val >= val else 0
        if root.left:
            result += self.dfs(root.left, max(root.val, val))
        if root.right:
            result += self.dfs(root.right, max(root.val, val))
        return result
        
```