<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 _rob(TreeNode* node, unordered_map<TreeNode*, int> &memo) {
        // Base cases
        if (!node) {
            return 0;
        }
        if (!(node->left || node->right)) {
            return node->val; 
        }

        // Already found
        auto it = memo.find(node);
        if (it != memo.end()) { 
            return it->second;
        }

        // Subproblems
        int grandchildrenVal = 0; 
        grandchildrenVal += node->left ? _rob(node->left->left, memo) : 0; 
        grandchildrenVal += node->left ? _rob(node->left->right, memo) : 0; 
        grandchildrenVal += node->right ? _rob(node->right->left, memo) : 0; 
        grandchildrenVal += node->right ? _rob(node->right->right, memo) : 0; 

        int useSelf = node->val + grandchildrenVal;
        int skipSelf = _rob(node->left, memo) + _rob(node->right, memo); 
        int optimum = max(useSelf, skipSelf); 

        memo.insert({node, optimum}); 
        return optimum;
    }
public:
    int rob(TreeNode* root) {
        // Recursive pull DP
        // Use hashMap for memoization: TreeNode* -> payoff
        // Subproblem: maximum payoff attainable in subtree T
        // Relate: dp[T] = max(dp[rootT left Subtree] + dp[rootT right Subtree], 
        // cur + sum(dp[rootT left grandchild trees]) + sum(dp[rootT right grandchild trees]))
        // Topological order: From leaf nodes to root node
        // Base case: Leaf nodes and null
        unordered_map<TreeNode*, int> memo;
        return _rob(root, memo);
    }
};
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