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+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)**
+
+    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:**
+
+**![](https://assets.leetcode.com/uploads/2020/04/02/test_sample_2.png)**
+
+    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]`.
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diff --git a/1544_count-good-nodes-in-binary-tree/python3/solution.py b/1544_count-good-nodes-in-binary-tree/python3/solution.py
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+# Time: O(N)
+# Space: O(N)
+
+# 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:
+        def dfs(node, max_value) -> int:
+            '''
+            Perform pre-order traversal and keep track of max
+            elements in the tree. Any subsequent traversal can
+            then compare against the updated max_value to see if
+            it's a good node or not
+            '''
+            
+            # If no left/right nodes, then we can just return 0
+            if not node: return 0
+            
+            # Current node is good if it's value is greater than or
+            # equal to the `max_value` seen so far
+            res = 1 if max_value <= node.val else 0
+            
+            # Compute the new max value, the current node could be it
+            max_value = max(max_value, node.val)
+            
+            # Do traversal on left and right nodes and add their units
+            res += dfs(node.left, max_value) + dfs(node.right, max_value)
+            
+            # This will indicate the count
+            return res
+        
+        
+        return dfs(root, root.val)