<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 dp[10005][2];
    map<int, int>mp;
    int count1 = 0;
    int DP(TreeNode* root, int node, int takenParent)
    {
        if(dp[node][takenParent]!=-1) return dp[node][takenParent];
        int val = 0, val1 = 0;
        if(!takenParent){
            val = mp[root->val];
            if(root->left != NULL) val += DP(root->left, root->left->val, 1);
            if(root->right != NULL) val += DP(root->right, root->right->val, 1);
        }
        if(root->left != NULL) val1 +=  DP(root->left, root->left->val, 0);
        if(root->right != NULL) val1 += DP(root->right, root->right->val, 0);
        return dp[node][takenParent] = max(val, val1);
    }
    void DFS(TreeNode* root)
    {
        if(root==NULL) return;
        mp[count1] = root->val;
        root->val = count1;
        count1 += 1;
        DFS(root->left);
        DFS(root->right);
    }

    int rob(TreeNode* root) {
        DFS(root);
        memset(dp, -1, sizeof(dp));
        return DP(root, 0, 0);
    }
};
18908