<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 {
public:
    int rob(TreeNode* root) {
        int level = 0, ret = 0;
        queue<TreeNode*> q;
        unordered_map<int, vector<TreeNode*>> m;
        unordered_map<TreeNode*, int> dp;
        q.push(root);
        while(!q.empty()){
            int n = q.size();
            for(int i=0; i<n; i++){
                TreeNode* tmp = q.front();
                q.pop();
                m[level].push_back(tmp);
                if(tmp->left){
                    q.push(tmp->left);
                }
                if(tmp->right){
                    q.push(tmp->right);
                }
            }
            level++;
        }
        for(int i=level-1; i>=0; i--){
            for(auto node:m[i]){
                int tmp1 = 0, tmp2 = node->val;
                if(node->left){
                    tmp1 += dp[node->left];
                    if(node->left->left){
                        tmp2 += dp[node->left->left];
                    }
                    if(node->left->right){
                        tmp2 += dp[node->left->right];
                    }
                }
                if(node->right){
                    tmp1 += dp[node->right];
                    if(node->right->left){
                        tmp2 += dp[node->right->left];
                    }
                    if(node->right->right){
                        tmp2 += dp[node->right->right];
                    }
                }
                dp[node] = max(tmp1, tmp2);
            }
        }
        return dp[root];
    }
};
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