From 987eb7ab5b0da7d4d27a0cc8d01df56d96c1a081 Mon Sep 17 00:00:00 2001
From: Prad Nukala <prad@sonr.io>
Date: Fri, 4 Sep 2026 16:44:07 -0400
Subject: [PATCH] feat(work): add solution for LeetCode 1448 Count Good Nodes
 in Binary Tree

---
 .../1448.count-good-nodes-in-binary-tree.py   | 94 +++++++++++++++++++
 1 file changed, 94 insertions(+)
 create mode 100644 work/1/Medium/Tree/1448.count-good-nodes-in-binary-tree.py

diff --git a/work/1/Medium/Tree/1448.count-good-nodes-in-binary-tree.py b/work/1/Medium/Tree/1448.count-good-nodes-in-binary-tree.py
new file mode 100644
index 0000000..1ee666a
--- /dev/null
+++ b/work/1/Medium/Tree/1448.count-good-nodes-in-binary-tree.py
@@ -0,0 +1,94 @@
+"""
+1448. Count Good Nodes in Binary Tree
+Difficulty: Medium
+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].
+"""
+
+
+# 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_val):
+            if not node:
+                return 0
+
+            is_good = 1 if node.val >= max_val else 0
+            new_max = max(max_val, node.val)
+
+            return is_good + dfs(node.left, new_max) + dfs(node.right, new_max)
+
+        return dfs(root, root.val)