<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 minCameraCover(TreeNode* root) {
        int cnt= 0;
        pair<bool, bool>f=  sol(root, cnt);
        if(!f.first && !f.second) cnt++;
        return cnt;

    }
    pair<bool, bool> sol(TreeNode* root, int &cnt){
        if(!root || (!root->left && !root->right)) return {false, false};
        bool flag= 0, cover= 0;
        if(root->left){
            if(!root->left->left && !root->left->right){
                flag = true;
            }
            else{
                pair<bool, bool>f1 = sol(root->left, cnt);
                if(!f1.first && !f1.second) flag= 1;
                if(f1.first) cover = 1;
            }
        }
        if(root->right){
            if(!root->right->left && !root->right->right){
                flag = true;
            }
            else{
                pair<bool, bool>f1 = sol(root->right, cnt);
                if(!f1.first && !f1.second) flag= 1;
                if(f1.first) cover = 1;
            }
        }
        if(flag) {cnt++; cover= 1;}
        return {flag, cover};

    }
};
20200