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)