<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>

/**
 * 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 {
        int check(TreeNode* root,bool flag,unordered_map <TreeNode*,pair<int,int>> &ourmap){
                 if(root == NULL){
                    return 0;
                 }
                 int a = 0;
                 if(flag){
                    a = 1;
                 }      
                 if(ourmap.count(root) > 0){
                    if(a == 0){
                       if(ourmap[root].first != -1){
                          return ourmap[root].first;
                       }
                    }
                    if(a == 1){
                       if(ourmap[root].second != -1){
                          return ourmap[root].second;
                       }  
                    }
                 }

                 int res1 = 0;
                 int res2 = 0;

                 if(flag){
                    res1 = check(root->left,flag,ourmap) + check(root->right,flag,ourmap);
                    res2 = root->val + check(root->left,false,ourmap) + check(root->right,false,ourmap); 
                 }
                 else{
                    res1 = check(root->left,true,ourmap) + check(root->right,true,ourmap);
                 }
                 int res = max(res1,res2); 
                 if(ourmap.count(root) > 0){
                    pair<int,int> p = ourmap[root];
                    if(a == 0){
                       p.first = res;
                    }
                    else{
                       p.second = res;
                    }
                 } 
                 else{
                    pair<int,int> p;
                    if(a == 0){
                       p.first = res;
                       p.second = -1;
                    }
                    else{
                       p.first = -1; 
                       p.second = res;
                    }
                    pair <TreeNode*,pair<int,int>> p1(root,p);  
                    ourmap.insert(p1);        
                 } 

                 return res;      
        } 

public:
    int rob(TreeNode* root) { 
            unordered_map <TreeNode*,pair<int,int>> ourmap;
            int result = check(root,true,ourmap);  

            return result;  
    }
};


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