<p>You are given the <code>root</code> 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.</p>

<p>Return <em>the minimum number of cameras needed to monitor all nodes of the tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png" style="width: 138px; height: 163px;" />
<pre>
<strong>Input:</strong> root = [0,0,null,0,0]
<strong>Output:</strong> 1
<strong>Explanation:</strong> One camera is enough to monitor all nodes if placed as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png" style="width: 139px; height: 312px;" />
<pre>
<strong>Input:</strong> root = [0,0,null,0,null,0,null,null,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>Node.val == 0</code></li>
</ul>

/**
 * 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 {
    bool isLeaf(TreeNode* n) {
        return !n->left && !n->right;
    }

    bool isCamera(TreeNode* n) {
        return n && n->val == CAM;
    }

    bool isMonitored(TreeNode* n) {
        return n && n->val == MON;
    }

    bool isUnmon(TreeNode* n) {
        return n && n->val == 0;
    };

    const int CAM = 2;
    const int MON = 1;
public:
    int minCameraCover(TreeNode* root) {
        if(!root->left && !root->right)
            return 1;

        queue<TreeNode*> q;
        vector<TreeNode*> v;   

        q.push(root);
        while(!q.empty()) {
            TreeNode* n = q.front(); 
            q.pop();
            v.push_back(n);

            if(n->left)
                q.push(n->left);
            if(n->right)
                q.push(n->right);
        }

        reverse(v.begin(), v.end()); 

        // now `v` has an inverse layered repr of a tree: from leaves to the root

        int numCameras = 0;
        for(TreeNode* n: v) {
            // place cameras as high in a tree as possible
            if(isLeaf(n))
                continue;
            if(isUnmon(n->left) || isUnmon(n->right)) {
                numCameras++;
                n->val = CAM;
                continue;
            }
            if(isCamera(n->left) || isCamera(n->right) ) {
                n->val = MON;
                continue;
            }
            if(n == root && !isCamera(n->left) && !isCamera(n->right)) {
                numCameras++;
                n->val = CAM;
                continue;
            }
        }
    
        return numCameras;
    }

};
21000