LC 968. Binary Tree Cameras

 

Given a binary tree, we install cameras on the nodes of the tree. 

Each camera at a node can monitor its parent, itself, and its immediate children.

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

 

Example 1:

 

 

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




Example 2:

Input: [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. 

Note:

1. The number of nodes in the given tree will be in the range [1, 1000].
2. Every node has value 0.

 

作了好久，仍是写不出来，看了网上的答案。确实很精妙。利用了返回值表达了三种状态，利用引用存储最后结果。

还有一点就是在放相机的时候，在递归函数里，父节点必定比节点有优点，能放父节点必定放父节点，拿叶节点来

举例，此时若是放叶节点，能影响到的点只有它自己和父节点，而父节点能影响到子节点，自己，和它的父节点。

因此这是一个贪心算法。

 

class Solution { public: int minCameraCover(TreeNode* root) { int sum=0; if(dfs(root,sum)==0)   sum++;// if root is not monitored, we place an additional camera here
        return sum; } int dfs(TreeNode * tr, int& sum){ if(!tr) return 1; int l=dfs(tr->left,sum), r=dfs(tr->right,sum); if(l==0||r==0){// if at least 1 child is not monitored, we need to place a camera at current node 
            sum++; return 2; }else if(l==2||r==2){// if at least 1 child has camera, the current node if monitored. Thus, we don't need to place a camera here 
            return 1; }else{// if both children are monitored but have no camera, we don't need to place a camera here. We place the camera at its parent node at the higher level. 
            return 0; } return -1;// this return statement won't be triggered
 } };