From ed4543b30042ab127a261b660054e54aaaf3102f Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Thu, 1 Jun 2023 04:03:00 -0500
Subject: [PATCH] =?UTF-8?q?Update=200968.=E7=9B=91=E6=8E=A7=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=A0=91.md?=
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---
 problems/0968.监控二叉树.md | 114 ++++++++++++++++++++++---------
 1 file changed, 81 insertions(+), 33 deletions(-)

diff --git a/problems/0968.监控二叉树.md b/problems/0968.监控二叉树.md
index fdec56ca..e2ba8ebf 100644
--- a/problems/0968.监控二叉树.md
+++ b/problems/0968.监控二叉树.md
@@ -371,56 +371,104 @@ class Solution {
 
 
 ### Python
-
+贪心（版本一）
 ```python
 class Solution:
-    def minCameraCover(self, root: TreeNode) -> int:
-        # Greedy Algo:
+         # 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
+    def minCameraCover(self, root: TreeNode) -> int:
+        # 定义递归函数
+        result = [0]  # 用于记录摄像头的安装数量
+        if self.traversal(root, result) == 0:
+            result[0] += 1
 
-        result = 0
-        # 从下往上遍历：后序（左右中）
-        def traversal(curr: TreeNode) -> int:
-            nonlocal result
+        return result[0]
 
-            if not curr: return 2
-            left = traversal(curr.left)
-            right = traversal(curr.right)
+        
+    def traversal(self, cur: TreeNode, result: List[int]) -> int:
+        if not cur:
+            return 2
 
-            # Case 1:
-            # 左右节点都有覆盖
-            if left == 2 and right == 2:
-                return 0
+        left = self.traversal(cur.left, result)
+        right = self.traversal(cur.right, result)
 
-            # 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
+        # 情况1: 左右节点都有覆盖
+        if left == 2 and right == 2:
+            return 0
 
-            # 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
+        # 情况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 左节点无覆盖，右节点有摄像头
+        # 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
+        if left == 0 or right == 0:
+            result[0] += 1
+            return 1
 
-            # 其他情况前段代码均已覆盖
+        # 情况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
+        if left == 1 or right == 1:
+            return 2
 
-        if traversal(root) == 0:
-            result += 1
 
-        return result
 ```
+贪心（版本二）利用elif精简代码
+```python
+class Solution:
+         # 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
+    def minCameraCover(self, root: TreeNode) -> int:
+        # 定义递归函数
+        result = [0]  # 用于记录摄像头的安装数量
+        if self.traversal(root, result) == 0:
+            result[0] += 1
 
+        return result[0]
+
+        
+    def traversal(self, cur: TreeNode, result: List[int]) -> int:
+        if not cur:
+            return 2
+
+        left = self.traversal(cur.left, result)
+        right = self.traversal(cur.right, result)
+
+        # 情况1: 左右节点都有覆盖
+        if left == 2 and right == 2:
+            return 0
+
+        # 情况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 左节点无覆盖，右节点有摄像头
+        # 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[0] += 1
+            return 1
+
+        # 情况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
+        else:
+            return 2
+
+
+
+
+```
 ### Go
 
 ```go