<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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rob(self, root: Optional[TreeNode]) -> int:
        
        def recur(root, name, memo):
            if root == None:
                return 0
            if name in memo:
                return memo[name]
            left_side = recur(root.left.left, name * 4, memo) + recur(root.left.right, name * 4 + 1, memo) if root.left != None else 0
            right_side = recur(root.right.left, name * 4 + 2, memo) + recur(root.right.right, name * 4 + 3, memo) if root.right != None else 0
            # use_this = left_side + right_side + root.val
            # skip_this = recur(root.left, name * 2, memo) + recur(root.right, name * 2 + 1, memo)
            memo[name] = max(left_side + right_side + root.val, recur(root.left, name * 2, memo) + recur(root.right, name * 2 + 1, memo))
            return memo[name]
        return recur(root, 1, dict())
18900