# [Solved] 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. ## Question

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 = 
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]`.

## Python Solution

```# 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

count = 0
def traversal(root,mx):
global count
if root==None:return
mx = max(root.val,mx)
if root.val>=mx:
count+=1
traversal(root.left,mx)
traversal(root.right,mx)

class Solution:
def goodNodes(self, root: TreeNode) -> int:
global count
count = 0
mx = root.val
traversal(root,mx)
return count```