一、题目

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.

Example 1:

Input: root = [0,0,null,0,0]
Output: 1
Explanation: One camera is enough to monitor all nodes if placed as shown.
Example 2:

Input: root = [0,0,null,0,null,0,null,null,0]
Output: 2
Explanation: At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.

Constraints:

The number of nodes in the tree is in the range [1, 1000].
Node.val == 0

二、题解

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:int res = 0;/*三种状态0-无覆盖1-有摄像头2-有覆盖*/int traversal(TreeNode* cur){//空节点：默认有覆盖if(cur == nullptr) return 2;int left = traversal(cur->left);int right = traversal(cur->right);//左右节点均有覆盖if(left == 2 && right == 2) return 0;//左节点无覆盖或右节点无覆盖if(left == 0 || right == 0){res++;return 1;}//左右节点中至少一个有摄像头if(left == 1 || right == 1) return 2;return -1;}int minCameraCover(TreeNode* root) {//如果根节点无覆盖if(traversal(root) == 0) res++;return res;}
};