<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 {
map<pair<TreeNode*,pair<int,int>>,int>mp;
int solve(TreeNode* root, int cur, int par) {
    if (!root){
        return cur == 1 ? INT_MAX / 2 : 0;
    }
    if(!root->left && !root->right){
        if(cur==1)
            return 1;
        if(cur==0 && par==0){
            return INT_MAX/2;
        }
        if(cur==0 && par==1){
            return 0;
        }
    }
    if(mp.find({root,{cur,par}})!=mp.end()) {
        return mp[{root, {cur,par}}];
    }
    int res;
    if (cur == 1) {
        res = 1 + min(solve(root->left, 0,1), solve(root->left, 1,1)) +
                    min(solve(root->right, 0,1), solve(root->right, 1,1));
    }else{
        if(par == 0){
        res=min(solve(root->left, 1 , 0) + min(solve(root->right, 0, 0),solve(root->right,1,0)),
                solve(root->right, 1, 0) + min(solve(root->left, 0, 0),solve(root->left,1,0)));
        }else{
            res = min(solve(root->left,1,0),solve(root->left,0,0)) + 
                  min(solve(root->right,1,0),solve(root->right,0,0));  
        }
    }
    return mp[{root, {cur,par}}] = res;
}
public:
    int minCameraCover(TreeNode* root) {
        mp.clear();
        int ans = min(solve(root,1,0),solve(root,0,0));
        return ans;
    }
};
28000