<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:
        
        # Memo
        
        # Dynamic programming function
        def foo(x):

            # Check if th tree is empty
            if not x:
                # Return the tuple 0,0
                return (0, 0)
            
            # If the left and right is none
            if not x.left and not x.right: 
                
                # Return the value of x and 0
                return (x.val, 0)
            
            # Return two possibilities
            
            # 1. Left node is robbed and the left node is not robbed
            lrobbed, lnotrobbed = foo(x.left)

            # 2. right node is robbed and the right node is not robbed
            rrobbed, rnotrobbed = foo(x.right)
            
            # After knowing the information above check the amount
            # It is the value of x added by the amount of l and r are not robbed 
            check = x.val + lnotrobbed + rnotrobbed
            
            # Compare the max of left robbed condition, and righ robbed condition
            # Add the two max values together
            check_max = max(lrobbed, lnotrobbed) + max(rrobbed, rnotrobbed)

            # Return two values, first the value of the value where the robbed is stop
            # second is the value when there are opportunieties that left or right robbed
            return (check, check_max)
        
        # Return the answer which is the max amount of money the thief can rob
        return max(foo(root))
        
18000