# LeetCode — Count Good Nodes in Binary Trees

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]`

.

Solution:

Using depth-first search to find out all the nodes greater than X, if the nodes are greater, then add it into numOfGoodNodes.