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].
```

- code

```
class Solution {
private int result = 0;
public int goodNodes(TreeNode root) {
dfs(root, Integer.MIN_VALUE);
return result;
}
private void dfs(TreeNode node, int curMax){
if (node == null) return;
if (node.val >= curMax) result++;
dfs(node.left, Math.max(curMax, node.val));
dfs(node.right, Math.max(curMax, node.val));
}
}
```