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.

Solution: Greedy + Recursion

Time complexity: O(n)
Space complexity: O(h)

enum class State { NONE = 0, COVERED = 1, CAMERA = 2 };

class Solution {
public:
  int minCameraCover(TreeNode* root) {        
    if (dfs(root) == State::NONE) ++ans_;
    return ans_;
  }
private:  
  int ans_ = 0;
  State dfs(TreeNode* root) {
    if (!root) return State::COVERED;
    State l = dfs(root->left);
    State r = dfs(root->right);
    if (l == State::NONE || r == State::NONE) {
      ++ans_;
      return State::CAMERA;
    }
    if (l == State::CAMERA || r == State::CAMERA)
      return State::COVERED;    
    return State::NONE;
  }
};

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