<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:
    int solve(TreeNode *root, unordered_map<TreeNode *,TreeNode *>&mp,int &res){
     if(root==NULL){
        return 1;
     }
     if(root->left==NULL && root->right==NULL){
        return 0;
     }
     if(root->left){
     mp[root->left]=root;}
     if(root->right){
     mp[root->right]=root;}
     int x=solve(root->left,mp,res);
     int y=solve(root->right,mp,res);
     int g=min(x,y);
     if(g==0){
        res++;
        root->val=1;
        if(mp[root]!=NULL){
        mp[root]->val=1;}
        return 1;
     }
     if(root->val==0){
return 0;
     }
return 1;
    }
    int minCameraCover(TreeNode* root) {
       int res=0;
       if(root==NULL){
        return 0;
       }
     
       unordered_map<TreeNode *,TreeNode *>mp;
       solve(root,mp,res); 
       if(root->val==0){
        res=res+1;
       }
       return res;
    }
};
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