目录

一，题目描述

英文描述

中文描述

示例与说明

二，解题思路

算法

状态划分 

分情况讨论

三，AC代码

C++

Java

四，解题过程

第一博

第二搏

第三博

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一，题目描述

英文描述

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.

Return the minimum number of cameras needed to monitor all nodes of the tree.

中文描述

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。

示例与说明

提示：

1. 给定树的节点数的范围是 [1, 1000]。
2. 每个节点的值都是 0。

来源：力扣（LeetCode）
链接：力扣
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

二，解题思路

参考@笨猪爆破组【「手画图解」从递归优化到树形DP | 监控二叉树】，文章介绍的比较详细，感兴趣的可以戳链接。

算法

状态划分 

对于一个非空节点可以分为三种状态：

1. 安装摄像头withCam（被自己监控）；
2. 不安装摄像头（被父节点监控watchByFat）；
3. 不安装摄像头（被子节点监控watchBySon）；

若节点为空，需要注意：

• 空节点不可能放置摄像头（此情况可以将返回值设置为INT_MAX，从而避免错误解被纳入结果考虑范围）；
• 空节点被父节点监控或是被子节点监控都返回0即可；

分情况讨论

1，当前节点安装摄像头：（子节点可装可不装，都会被当前节点监控）

• minNum = min(minNum, left["withCam"] + right["watchByFat"]);
• minNum = min(minNum, left["watchByFat"] + right["withCam"]);
• minNum = min(minNum, left["watchByFat"] + right["watchByFat"]);
• minNum = min(minNum, left["withCam"] + right["withCam"]);（此情况可以不用考虑，因为当前节点及左右子节点都安装摄像头的策略，基本不会是最优策略）

2，当前节点不安装摄像头：（子节点不可能被当前节点监控。只能被自己监控，或者被他们的子节点监控）

（当前节点被父节点监控。子节点可装可不装）

• minNum = min(minNum, left["withCam"] + right["withCam"]);
• minNum = min(minNum, left["withCam"] + right["watchBySon"]);
• minNum = min(minNum, left["watchBySon"] + right["withCam"]);
• minNum = min(minNum, left["watchBySon"] + right["watchBySon"]);

（当前节点被子节点监控。子节点至少有一个摄像头）

• minNum = min(minNum, left["withCam"] + right["withCam"]);
• minNum = min(minNum, left["withCam"] + right["watchBySon"]);
• minNum = min(minNum, left["watchBySon"] + right["withCam"]);

至此便考虑到了所有可能成为解的情况，并将计算的结果一步步返回递归的上一层，从而重复利用。

三，AC代码

C++

class Solution {
public:int minCameraCover(TreeNode* root) {unordered_map<string, int> ans = getMin(root);int minNum = 99999999;minNum = min(minNum, ans["withCam"]);// minNum = min(minNum, ans["watchByFat"]); // !!!根节点怎么可能被父节点监控minNum = min(minNum, ans["watchBySon"]);return minNum;}unordered_map<string, int> getMin(TreeNode* root) {unordered_map<string, int> ans;if (root == nullptr) {ans["withCam"] = 99999999;// 空节点不可能有摄像头，所以这里返回一个足够大的值，使得这种情况不可能被纳入最优解考虑范围ans["watchByFat"] = 0;// 空节点不需要被监控ans["watchBySon"] = 0;// 空节点不可能被子节点监控return ans;// !!!必须在这里返回}unordered_map<string, int> left = getMin(root->left);unordered_map<string, int> right = getMin(root->right);// 当前节点安装摄像头int minNum = 99999999;minNum = min(minNum, left["withCam"] + right["watchByFat"]);minNum = min(minNum, left["watchByFat"] + right["withCam"]);minNum = min(minNum, left["watchByFat"] + right["watchByFat"]);int withCam = 1 + minNum;// 当前节点不安装摄像头// 被父节点监控minNum = 99999999;minNum = min(minNum, left["withCam"] + right["withCam"]);minNum = min(minNum, left["withCam"] + right["watchBySon"]);minNum = min(minNum, left["watchBySon"] + right["withCam"]);minNum = min(minNum, left["watchBySon"] + right["watchBySon"]);int watchByFat = minNum;// 被子节点监控（子节点至少有一个摄像头）minNum = 99999999;minNum = min(minNum, left["withCam"] + right["withCam"]);minNum = min(minNum, left["withCam"] + right["watchBySon"]);minNum = min(minNum, left["watchBySon"] + right["withCam"]);int watchBySon = minNum;ans["withCam"] = withCam;ans["watchByFat"] = watchByFat;ans["watchBySon"] = watchBySon;return ans;}};

Java

class Solution {public int minCameraCover(TreeNode root) {Reult ans = getMin(root);return Math.min(ans.withCam, ans.watchBySon);}public Reult getMin(TreeNode root) {Reult ans = new Reult();if (root == null) {ans.withCam = 99999999;ans.watchByFat = 0;ans.watchBySon = 0;return ans;}Reult left = getMin(root.left);Reult right = getMin(root.right);// 在该节点放置摄像头ans.withCam = 99999999;ans.withCam = Math.min(ans.withCam, left.withCam + right.watchByFat);ans.withCam = Math.min(ans.withCam, left.watchByFat + right.withCam);ans.withCam = Math.min(ans.withCam, left.watchByFat + right.watchByFat);ans.withCam++;// 该节点不放置摄像头（被父节点监控）ans.watchByFat = 99999999;ans.watchByFat = Math.min(ans.watchByFat, left.withCam + right.watchBySon);ans.watchByFat = Math.min(ans.watchByFat, left.watchBySon + right.withCam);ans.watchByFat = Math.min(ans.watchByFat, left.withCam + right.withCam);ans.watchByFat = Math.min(ans.watchByFat, left.watchBySon + right.watchBySon);// 该节点不放置摄像头（被子节点监控）ans.watchBySon = 99999999;ans.watchBySon = Math.min(ans.watchBySon, left.withCam + right.watchBySon);ans.watchBySon = Math.min(ans.watchBySon, left.watchBySon + right.withCam);ans.watchBySon = Math.min(ans.watchBySon, left.withCam + right.withCam);return ans;}
}
class Reult {int withCam;int watchByFat;int watchBySon;
}

 

四，解题过程

第一博

这种一般直接想到用递归，只要归纳全部的情况就能得到正确答案，虽然看上去效率很低，但是现实现出来再说

class Solution {public int minCameraCover(TreeNode root) {return Math.min(getAns(root, 0), getAns(root, 1));}// flag表示当前节点是否安装摄像头public int getAns(TreeNode root, int flag) {if (root == null) return 0;int minNum = 99999999;if (flag == 1) {minNum = Math.min(minNum, getAns(root.left, 1) + getAns(root.right, 1));minNum = Math.min(minNum, getAns(root.left, 1) + getAns(root.right, 0));minNum = Math.min(minNum, getAns(root.left, 0) + getAns(root.right, 1));minNum = Math.min(minNum, getAns(root.left, 0) + getAns(root.right, 0));} else {if (root.left != null && root.right != null) {minNum = Math.min(minNum, getAns(root.left, 1) + getAns(root.right, 1));minNum = Math.min(minNum, getAns(root.left, 1) + getAns(root.right, 0));minNum = Math.min(minNum, getAns(root.left, 0) + getAns(root.right, 1));} else if (root.left == null && root.right == null) {return flag;} else if (root.left == null) {minNum = Math.min(minNum, getAns(root.left, 0) + getAns(root.right, 1));} else if (root.right == null) {minNum = Math.min(minNum, getAns(root.left, 1) + getAns(root.right, 0));}}System.out.println(minNum + flag);return minNum + flag;}
}

显然上面的方法没有考虑到父节点的状态，因此没有包含所有的状态，最后只过了121/171

第二搏

参考了大佬的解题思路，发现是自己考虑的情况不够全面 ，比如只考虑当前节点放不放摄像头，未考虑父节点或子节点对当前节点监控状态的影响，这样就无法完全保证覆盖关系，无法判断连续两个节点不放置摄像头的情况。

因此需要引入当前节点是否被监视的参数。

class Solution {public int minCameraCover(TreeNode root) {return Math.min(getNum(root, true, true), getNum(root, false, false));}// hasCame: 该节点是否有摄像头，isWatched: 是否被自己或父节点监控public int getNum(TreeNode root,boolean hasCame, boolean isWatched) {if (root == null) {if (hasCame) return 99999999;else return 0;}int minNum = 99999999;// 该节点有摄像头(子结点可装可不装，舍弃子结点全部安装的情况，因为该情况不可能作为最终解)if (hasCame) {minNum = Math.min(minNum, getNum(root.left, true, true) + getNum(root.right, false, true) + 1);minNum = Math.min(minNum, getNum(root.left, false, true) + getNum(root.right, true, true) + 1);minNum = Math.min(minNum, getNum(root.left, false, true) + getNum(root.right, false, true) + 1);} // 该节点无摄像头else {// 父节点有摄像头，被父节点监控(子结点可装可不装)if (isWatched) {minNum = Math.min(minNum, getNum(root.left, true, true) + getNum(root.right, true, true));minNum = Math.min(minNum, getNum(root.left, false, false) + getNum(root.right, true, true));minNum = Math.min(minNum, getNum(root.left, true, true) + getNum(root.right, false, false));minNum = Math.min(minNum, getNum(root.left, false, false) + getNum(root.right, false, false));} // 父节点无摄像头，被子结点监控(至少有一个子结点要装摄像头)else {minNum = Math.min(minNum, getNum(root.left, true, true) + getNum(root.right, false, false));minNum = Math.min(minNum, getNum(root.left, false, false) + getNum(root.right, true, true));minNum = Math.min(minNum, getNum(root.left, true, true) + getNum(root.right, true, true));}}return minNum;}
}

显然，后面的测试用例都是超时了。

第三博

可以从上面的算法中看出，为保证正确性，多次对同一颗子树进行求解，导致计算重复，如果能够在一次递归中将各种情况的计算结果一次性返回，并且重复利用就可以提高效率。这便是树形DP

解题过程用的是map，这样看起来更直观一点，但是开销比较大，想提高效率可以用数组代替