337.md

337. House Robber III

Solution 1 (O(n))

# Dynamic programming
# 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
        """
        from collections import defaultdict
        if root is None:
            return 0
        self.dp1 = defaultdict(int)  # rob current point
        self.dp2 = defaultdict(int)  # do not rob current point
        self.dfs(root)
        return max(self.dp1[root], self.dp2[root])
    
    def dfs(self, root):
        if root.left is not None:
            self.dfs(root.left)
        if root.right is not None:
            self.dfs(root.right)
        self.dp1[root] = root.val + self.dp2[root.left] + self.dp2[root.right]
        self.dp2[root] = (max(self.dp1[root.left], self.dp2[root.left])
                          + max(self.dp1[root.right], self.dp2[root.right]))