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.

做了很久，還是寫不出來，看了網上的答案。確實很精妙。利用了返回值表達了三種狀態，利用引用儲存最後結果。

還有一點就是在放相機的時候，在遞迴函式裡，父節點一定比節點有優勢，能放父節點一定放父節點，拿葉節點來

舉例，此時如果放葉節點，能影響到的點只有它本身和父節點，而父節點能影響到子節點，本身，和它的父節點。

所以這是一個貪心演算法。

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
    }
};