From c8ddb46c9d4a000fbfb1498c0be32af758b911e5 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Tue, 19 Oct 2021 15:18:20 +0800
Subject: [PATCH] =?UTF-8?q?968:=E7=9B=91=E6=8E=A7=E4=BA=8C=E5=8F=89?=
 =?UTF-8?q?=E6=A0=91?=
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---
 .../leetcode/editor/cn/BinaryTreeCameras.java | 86 +++++++++++++++++++
 .../cn/doc/content/BinaryTreeCameras.md       | 34 ++++++++
 2 files changed, 120 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/BinaryTreeCameras.java
 create mode 100644 src/main/java/leetcode/editor/cn/doc/content/BinaryTreeCameras.md

diff --git a/src/main/java/leetcode/editor/cn/BinaryTreeCameras.java b/src/main/java/leetcode/editor/cn/BinaryTreeCameras.java
new file mode 100644
index 0000000..3d62486
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/BinaryTreeCameras.java
@@ -0,0 +1,86 @@
+//给定一个二叉树，我们在树的节点上安装摄像头。 
+//
+// 节点上的每个摄影头都可以监视其父对象、自身及其直接子对象。 
+//
+// 计算监控树的所有节点所需的最小摄像头数量。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//
+// 输入：[0,0,null,0,0]
+//输出：1
+//解释：如图所示，一台摄像头足以监控所有节点。
+// 
+//
+// 示例 2： 
+//
+// 
+//
+// 输入：[0,0,null,0,null,0,null,null,0]
+//输出：2
+//解释：需要至少两个摄像头来监视树的所有节点。 上图显示了摄像头放置的有效位置之一。
+// 
+//
+// 
+//提示： 
+//
+// 
+// 给定树的节点数的范围是 [1, 1000]。 
+// 每个节点的值都是 0。 
+// 
+// Related Topics 树 深度优先搜索 动态规划 二叉树 👍 330 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.TreeNode;
+
+//968:监控二叉树
+class BinaryTreeCameras {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new BinaryTreeCameras().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+
+    /**
+     * Definition for a binary tree node.
+     * public class TreeNode {
+     * int val;
+     * TreeNode left;
+     * TreeNode right;
+     * TreeNode() {}
+     * TreeNode(int val) { this.val = val; }
+     * TreeNode(int val, TreeNode left, TreeNode right) {
+     * this.val = val;
+     * this.left = left;
+     * this.right = right;
+     * }
+     * }
+     */
+    class Solution {
+        public int minCameraCover(TreeNode root) {
+            int[] array = dfs(root);
+            return array[1];
+        }
+
+        public int[] dfs(TreeNode root) {
+            if (root == null) {
+                return new int[]{Integer.MAX_VALUE / 2, 0, 0};
+            }
+            int[] leftArray = dfs(root.left);
+            int[] rightArray = dfs(root.right);
+            int[] array = new int[3];
+            array[0] = leftArray[2] + rightArray[2] + 1;
+            array[1] = Math.min(array[0], Math.min(leftArray[0] + rightArray[1], rightArray[0] + leftArray[1]));
+            array[2] = Math.min(array[0], leftArray[1] + rightArray[1]);
+            return array;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/doc/content/BinaryTreeCameras.md b/src/main/java/leetcode/editor/cn/doc/content/BinaryTreeCameras.md
new file mode 100644
index 0000000..a72e7df
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/BinaryTreeCameras.md
@@ -0,0 +1,34 @@
+<p>给定一个二叉树，我们在树的节点上安装摄像头。</p>
+
+<p>节点上的每个摄影头都可以监视<strong>其父对象、自身及其直接子对象。</strong></p>
+
+<p>计算监控树的所有节点所需的最小摄像头数量。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/bst_cameras_01.png" style="height: 163px; width: 138px;"></p>
+
+<pre><strong>输入：</strong>[0,0,null,0,0]
+<strong>输出：</strong>1
+<strong>解释：</strong>如图所示，一台摄像头足以监控所有节点。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/bst_cameras_02.png" style="height: 312px; width: 139px;"></p>
+
+<pre><strong>输入：</strong>[0,0,null,0,null,0,null,null,0]
+<strong>输出：</strong>2
+<strong>解释：</strong>需要至少两个摄像头来监视树的所有节点。 上图显示了摄像头放置的有效位置之一。
+</pre>
+
+<p><br>
+<strong>提示：</strong></p>
+
+<ol>
+	<li>给定树的节点数的范围是&nbsp;<code>[1, 1000]</code>。</li>
+	<li>每个节点的值都是 0。</li>
+</ol>
+<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>动态规划</li><li>二叉树</li></div></div><br><div><li>👍 330</li><li>👎 0</li></div>
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