<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 {
public:

    // intuition - dp, as there is two choices to place or not place a camera

    unordered_map<TreeNode*, int> dp;

    int cover(TreeNode* root){ // camera at this root

        // 4 cases

        int a = minCameras(root->left) + minCameras(root->right);

        int b = minCameras(root->left) + 
                (root->right != NULL ? minCameras(root->right->left) + minCameras(root->right->right) : 0);

        int c = minCameras(root->right) + 
                (root->left != NULL ? minCameras(root->left->left) + minCameras(root->left->right) : 0);

        int d = (root->left != NULL ? minCameras(root->left->left) + minCameras(root->left->right) : 0)
                + (root->right != NULL ? minCameras(root->right->left) + minCameras(root->right->right) : 0);

        return min(a, min(b, min(c, d)));
    }

    int minCameras(TreeNode* root){

        if(root == NULL) return 0;

        if(dp.find(root) != dp.end()) return dp[root];

        // placing a camera at the root

        int m = 1 + cover(root);

        // not placing a camera at the root
        // 2 cases left or right

        int n1 = INT_MAX;

        if(root->left != NULL){

            n1 = 1 + cover(root->left) + minCameras(root->right);
        }

        int n2 = INT_MAX;

        if(root->right != NULL){

            n2 = 1 + cover(root->right) + minCameras(root->left);
        }

        return dp[root] = min(m, min(n1, n2));
    }


    int minCameraCover(TreeNode* root) {
        
        return minCameras(root);
    }
};
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