递归与二叉树_leetcode230

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


# 即中序遍历的第K个节点
class Solution1(object):
 def kthSmallest(self, root, k):
 self.count=k
 self.res=0

 def core(root):
 if root :

 core(root.left)

 self.count=self.count-1
 if self.count==0:
 self.res=root.val

 core(root.right)

 core(root)
 return self.res



# 这个栈的中序遍历好不规范呀
class Solution2(object):
 def kthSmallest(self, root, k):
 stack=[]
 while True:
 if root:
 stack.append(root)
 root=root.left
 else:
 root=stack.pop()
 k=k-1
 if k==0:
 return root.val
 else:
 root=root.right



class Solution3(object):
 def kthSmallest(self, root, k):

 res = []
 self.midOrder(root,res)
 return res[k-1]



 def midOrder(self,root,res):

 if not root:
 return


 self.midOrder(root.left,res)
 res.append(root.val)
 self.midOrder(root.right,res)

 return res




# 启示： 下层的K值无法改变上层的k值,所以要用全局变量
class Solution4(object):
 def kthSmallest(self, root, k):


 return self.midOrder(root,k)


 def midOrder(self,root,k):

 if not root:
 return

 self.midOrder(root.left,k)

 k = k-1
 if k == 0:
 return root.val

 self.midOrder(root.right,k)


 def createTreee(self):
 root = TreeNode(3)
 root.left = TreeNode(1)
 root.right = TreeNode(4)
 root.left.right = TreeNode(2)

 return root


class Solution5(object):
 def kthSmallest(self, root, k):
 self.count = 0
 self.res = 0

 self.midOrder(root,k)

 return self.res


 def midOrder(self,root,k):

 if not root:
 return


 self.midOrder(root.left,k)

 self.count += 1
 if k == self.count:
 self.res = root.val
 return

 self.midOrder(root.right,k)



class Solution6(object):
 def kthSmallest(self, root, k):
 if not root:
 return root

 count = k
 stack = []

 stack.append(root.left)
 



s = Solution4()

root = s.createTreee()

s.kthSmallest(root,1)