<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> covered;
    int count;

    void Solve(TreeNode* root, TreeNode* parent) {
        if (root == NULL) {
            return;
        }

        // Perform post-order traversal
        Solve(root->left, root);//pehle hum saare nodes mai traverse krenge and will start from the leaf nodes
        Solve(root->right, root);

        // If the current node is uncovered, or if its children are uncovered,
        // place a camera at the parent node.
        if ((parent == NULL && covered.count(root) == 0) || 
            (root->left != NULL && covered.count(root->left) == 0) || 
            (root->right != NULL && covered.count(root->right) == 0)) {

            // Place a camera on the parent node
            covered[parent] = 1;
            covered[root] = 1;
            if (root->left!= NULL){ covered[root->left] = 1;}
            if (root->right!= NULL) {covered[root->right] = 1;}
            count++;
        }
    }

    int minCameraCover(TreeNode* root) {
        count = 0;
        covered[NULL] = 1;  // Handle the case where the parent is NULL (i.e., root)
        Solve(root, NULL);

        // After traversal, check if root itself is uncovered
        if (covered.count(root) == 0) {
            count++;
        }

        return count;
    }
};

23800