Update 0968.监控二叉树.md

完善python代码，补充comment理解

problems/0968.监控二叉树.md

@@ -347,24 +347,57 @@ class Solution {
 
 Python：
 ```python
+# 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 minCameraCover(self, root: TreeNode) -> int:
+        # Greedy Algo:
+        # 从下往上安装摄像头：跳过leaves这样安装数量最少，局部最优 -> 全局最优
+        # 先给leaves的父节点安装，然后每隔两层节点安装一个摄像头，直到Head
+        # 0: 该节点未覆盖
+        # 1: 该节点有摄像头
+        # 2: 该节点有覆盖
+        
         result = 0
-        def traversal(cur):
+        # 从下往上遍历：后序（左右中）
+        def traversal(curr: TreeNode) -> int:
             nonlocal result
-            if not cur:
+            
-                return 2
+            if not curr: return 2
-            left = traversal(cur.left)
+            left = traversal(curr.left)
-            right = traversal(cur.right)
+            right = traversal(curr.right)
+
+            # Case 1:
+            # 左右节点都有覆盖
             if left == 2 and right == 2: 
                 return 0
+
+            # Case 2:
+                # left == 0 && right == 0 左右节点无覆盖
+                # left == 1 && right == 0 左节点有摄像头，右节点无覆盖
+                # left == 0 && right == 1 左节点有无覆盖，右节点摄像头
+                # left == 0 && right == 2 左节点无覆盖，右节点覆盖
+                # left == 2 && right == 0 左节点覆盖，右节点无覆盖
             elif left == 0 or right == 0: 
                 result += 1
                 return 1
+
+            # Case 3:
+                # left == 1 && right == 2 左节点有摄像头，右节点有覆盖
+                # left == 2 && right == 1 左节点有覆盖，右节点有摄像头
+                # left == 1 && right == 1 左右节点都有摄像头
             elif left == 1 or right == 1:
                 return 2
-            else: return -1
+            
-        if traversal(root) == 0: result += 1
+            # 其他情况前段代码均已覆盖
+
+        if traversal(root) == 0:
+            result += 1
+            
         return result
 ```
 Go：