<p>The thief has found himself a new place for his thievery again. There is only one entrance to this area, called <code>root</code>.</p>

<p>Besides the <code>root</code>, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if <strong>two directly-linked houses were broken into on the same night</strong>.</p>

<p>Given the <code>root</code> of the binary tree, return <em>the maximum amount of money the thief can rob <strong>without alerting the police</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/10/rob1-tree.jpg" style="width: 277px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [3,2,3,null,3,null,1]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/10/rob2-tree.jpg" style="width: 357px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [3,4,5,1,3,null,1]
<strong>Output:</strong> 9
<strong>Explanation:</strong> Maximum amount of money the thief can rob = 4 + 5 = 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int fun(TreeNode* root,int x,map<pair<TreeNode*,int>,int>& dp){
        if(root==nullptr)
        return 0;
        if(root->left==nullptr && root->right==nullptr){
            if(x==0){
                return 0;
            }
            else
            return root->val;
        }
        if(dp.find({root,x})!=dp.end())
        return dp[{root,x}];
        int ans=0;
        if(x==0){
            ans=max(fun(root->left,0,dp),fun(root->left,1,dp))+max(fun(root->right,0,dp),fun(root->right,1,dp));
        }
        else{
            ans=root->val+fun(root->left,0,dp)+fun(root->right,0,dp);
        }
        return dp[{root,x}]=ans;
    }
    int rob(TreeNode* root) {
        if(root==nullptr)
        return 0;
        map<pair<TreeNode*,int>,int> dp;
        int withrobbery = root->val + fun(root->left,0,dp)+fun(root->right,0,dp);
        int withoutrobbery = max(fun(root->left,0,dp),fun(root->left,1,dp))+max(fun(root->right,0,dp),fun(root->right,1,dp));
        return max(withrobbery,withoutrobbery);
    }
};
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