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)