<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) {}
 * };
 */

#define tn TreeNode* 
#define inf (1<<20)
class Solution {
public:

    map<pair<tn, pair<bool, bool>>, int> dp; 
    int dfs(TreeNode* node,bool cam, bool parCam){

        if(!node){
            if(cam) return 1e9;

            return 0;
        }

        if(!node->left and !node->right){

            if(cam) return 1;

            if(parCam) return 0;

            return 1e9;
        }

        if(dp.find({node, {cam, parCam}}) != dp.end())
            return dp[{node, {cam, parCam}}];

        if(cam){

            int var1 = min(dfs(node->left,0,1),dfs(node->left,1,1));
            int var2 = min(dfs(node->right,0,1),dfs(node->right,1,1));

            return dp[{node, {cam, parCam}}]=1+var1+var2;

        }

        else if(parCam){
            int var1 = min(dfs(node->left,0,0),dfs(node->left,1,0));
            int var2 = min(dfs(node->right,0,0),dfs(node->right,1,0));

            return dp[{node, {cam, parCam}}]=var1+var2;
        }
        else{
            int var1 = dfs(node->right,1,0) + min(dfs(node->left,0,0),dfs(node->left,1,0));
            int var2 = dfs(node->left,1,0) + min(dfs(node->right,0,0),dfs(node->right,1,0));

            return dp[{node, {cam, parCam}}]=min(var1,var2);
        }
    }
    int minCameraCover(TreeNode* root) {
        if(!root) return 0;

        if(!root->left and !root->right) return 1;

        dp.clear();
        return min(dfs(root,1,0),dfs(root,0,0));
    }
};
29000