LeetCode in Python 337. House Robber III

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

Output: 7 
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

只能隔层偷，如果直连的两层被盗就报警。用map[root]表示从root出发能偷的最大数（可能包含root，也可能不包含root），分两种情况计算。用map存结果避免反复递归。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def rob(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        m = {}
        
        def memorized(root):
            if not root: return 0
            if m.has_key(root): return m[root]
            
            oneStep = memorized(root.left) + memorized(root.right)
            twoSteps = 0
            if root.left:
                twoSteps += (memorized(root.left.left) + memorized(root.left.right))
            if root.right:
                twoSteps += (memorized(root.right.left) + memorized(root.right.right))
            
            m[root] = max((root.val + twoSteps), oneStep)
            return m[root]
        
        return memorized(root)