<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>

class Solution {
public:
    unordered_map<TreeNode*, bool>um;//bydefault 0 se initialised h...
    int count;

    void solve(TreeNode*root, TreeNode*parent) {

        if (root == NULL) return;

        solve(root->left, root);
        solve(root->right, root);

        if ((parent==NULL and um.count(root) == 0) or um.count(root->left) == 0 or
                um.count(root->right) == 0) {
            um[parent] = 1;
            um[root] = 1;
            um[root->left] = 1;
            um[root->right] = 1;
            count++;
        }
    }

    int minCameraCover(TreeNode* root) {
        count = 0;
        um.insert({NULL, true});

        solve(root, NULL);
        return count;

    }
};
23900