[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 = [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].

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