动态规划_leetcode337
Posted AceKo
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#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以上是关于动态规划_leetcode337的主要内容，如果未能解决你的问题，请参考以下文章