动态规划_leetcode337

#coding=utf-8

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


# 递归
class Solution1(object):
    def rob(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """

        if not root :
            return 0

        return  self.tryRob(root)


    def tryRob(self,root):


        if not root:
            return 0

        if root and not root.left and not root.right:
            return root.val


        leftChild = root.left
        rightChild = root.right

        res = root.val


        if leftChild:
            res += self.tryRob(leftChild.left)
            res += self.tryRob(leftChild.right)
        if rightChild:
            res += self.tryRob(rightChild.left)
            res += self.tryRob(rightChild.right)


        res = max(res,self.tryRob(leftChild)+self.tryRob(rightChild))

        return res


    def createTree1(self):

        root = TreeNode(3)
        root.left = TreeNode(4)
        root.right = TreeNode(5)

        root.left.left = TreeNode(1)
        root.left.right = TreeNode(3)

        root.right.right = TreeNode(1)



        return root


    def createTree2(self):
        root = TreeNode(3)

        root.left = TreeNode(2)
        root.right = TreeNode(3)

        root.left.right = TreeNode(3)
        root.right.right = TreeNode(1)

        return root



# s =  Solution1()
#
# root = s.createTree2()
#
# print s.rob(root)


# 记忆化递归
class Solution2(object):
    def rob(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """

        if not root :
            return 0

        self.memo = {}


        return  self.tryRob(root)


    def tryRob(self,root):


        if self.memo.has_key(root):
            return self.memo[root]

        if not root:
            return 0

        if root and not root.left and not root.right:
            self.memo[root] = root.val
            return root.val

        else:

            res1 = self.tryRob(root.left) + self.tryRob(root.right)

            res2 = root.val

            if root.left:
                res2 += self.tryRob(root.left.left)
                res2 += self.tryRob(root.left.right)

            if root.right:
                res2 += self.tryRob(root.right.left)
                res2 += self.tryRob(root.right.right)


            res = max(res1,res2)
            self.memo[root] = res

            return res




    def createTree1(self):

        root = TreeNode(3)
        root.left = TreeNode(4)
        root.right = TreeNode(5)

        root.left.left = TreeNode(1)
        root.left.right = TreeNode(3)

        root.right.right = TreeNode(1)



        return root


    def createTree2(self):
        root = TreeNode(3)

        root.left = TreeNode(2)
        root.right = TreeNode(3)

        root.left.right = TreeNode(3)
        root.right.right = TreeNode(1)

        return root



# s = Solution2()
#
# root = s.createTree2()
#
# print s.rob(root)



# 刘宇波的代码1
# 递归版
class Solution3(object):
    def rob(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """


        return self.tryRob(root,True)


    def tryRob(self,root,include):

        if not root:
            return 0



        res = self.tryRob(root.left,True) + self.tryRob(root.right,True)

        if include:
            res = max(res,root.val + self.tryRob(root.left,False) + self.tryRob(root.right,False))



        return res

# 刘宇波的代码2# 这个include 技巧用的好呀, 不同的递归函数用一个参数就合并了class Solution4(object):    def rob(self, root):        """        :type root: TreeNode        :rtype: int        """        self.memo = {}        return self.tryRob(root,True)    def tryRob(self,root,include):        if not root:            return 0        if self.memo.has_key(root):            return self.memo[root]        res = self.tryRob(root.left,True) + self.tryRob(root.right,True)        if include:            res = max(res,root.val + self.tryRob(root.left,False) + self.tryRob(root.right,False))        self.memo[root] = res        return res# 刘宇波的代码3# 二元返回值分别代表包含该节点的最大值和不包含该节点的最大值class Solution5(object):    def rob(self, root):        """        :type root: TreeNode        :rtype: int        """        res =  self.tryRob(root)        return res[1]    # res[0] 不包含root 的最大值    # res[1] 包含root 或者不包含 root 的最大值    def tryRob(self,root):        if not root:            return [0,0]        resL = self.tryRob(root.left)        resR = self.tryRob(root.right)        res=[0,0]        res[0] = resL[1] + resR[1]        res[1] = max(res[0],root.val + resL[0] + resR[0])        return res

动态规划_leetcode337

原文：https://www.cnblogs.com/lux-ace/p/10546622.html