<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 dp[1005][4];
int solve(TreeNode* root,int dis){
    if(root==NULL)return 0;
    if(dis==-1){
        return min({1+solve(root->left,0)+solve(root->right,0),(root->left&&root->right?solve(root->left,2)+solve(root->right,1):(root->left?solve(root->left,2):100000000)),(root->left&&root->right?solve(root->right,2)+solve(root->left,1):(root->right?solve(root->right,2):100000000))});
    }
    if(dp[root->val][dis]!=-1)return dp[root->val][dis];
    if(dis==2){
        return dp[root->val][dis]=1+solve(root->left,0)+solve(root->right,0);
    }
    if(dis==1&&!root->left&&!root->right)return dp[root->val][dis]=1;
    if(root->left&&root->right&&dis==1){
        return dp[root->val][dis]=min({1+solve(root->left,0)+solve(root->right,0),solve(root->left,2)+solve(root->right,1),solve(root->left,1)+solve(root->right,2)});
    }
    return dp[root->val][dis]=min({1+solve(root->left,0)+solve(root->right,0),solve(root->left,dis+1)+solve(root->right,dis+1)});
}
    int minCameraCover(TreeNode* root) {
        if(root==NULL)return 0;
        if(root->left==NULL&&root->right==NULL)return 1;
        TreeNode* tem=root;
        queue<TreeNode*>q;
        q.push(tem);
        int cnt=0;
        while(!q.empty()){
            TreeNode* nod=q.front();
            q.pop();
            nod->val=cnt++;
            if(nod->left){
                q.push(nod->left);
            }
            if(nod->right){
                q.push(nod->right);
            }
        }
        memset(dp,-1,sizeof(dp));
        return solve(root,-1);
    }
};
20700