<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) {
        // get the parent map and children map
        using Node = TreeNode;
        unordered_map<Node*, Node*> pmap;
        unordered_map<Node*, vector<Node*>> cmap;
        vector<Node*> nodelist;
        vector<Node*> front{root};
        while(!front.empty()) {
            vector<Node*> next;
            for(auto p: front) {
                nodelist.push_back(p);
                if(p->left) {
                    next.push_back(p->left);
                    pmap[p->left] = p;
                    cmap[p].push_back(p->left);
                }
                if(p->right) {
                    next.push_back(p->right);
                    pmap[p->right] = p;
                    cmap[p].push_back(p->right);
                }
            }
            swap(next, front);
        }
        set<Node*> chosen;
        set<Node*> covered;

        while(!nodelist.empty()) {
            auto node = nodelist.back();
            if(covered.count(node)) {
                nodelist.pop_back();
                continue;
            }
            if(pmap.count(node)) {
                auto choice = pmap[node];
                chosen.insert(choice);
                covered.insert(choice);
                if(pmap.count(choice)) {
                    covered.insert(pmap[choice]);
                }
                for(auto child: cmap[choice]) {
                    covered.insert(child);
                }
                nodelist.pop_back();
            }
            else {
                chosen.insert(node);
                covered.insert(node);
                nodelist.pop_back();
            }
        }

        return chosen.size();

    }

};
25900