题解 | #二叉搜索树的第k个节点#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param proot TreeNode类 
# @param k int整型 
# @return int整型
#
class Solution:
    # 先定义返回和当前访问数量
    def __init__(self):
        # 记录结果
        self.res = None
        # 记录访问数量
        self.count = 0
    def zhongxubianli(self,proot,k):
        # 如果当前结点为空或者count大于k则返回
        if proot == None or self.count > k:
            return None
        # 中序遍历
        self.zhongxubianli(proot.left,k)
        self.count = self.count + 1
        if self.count == k:
            self.res = proot
        self.zhongxubianli(proot.right,k)

    def KthNode(self , proot: TreeNode, k: int) -> int:
        # write code here
        self.zhongxubianli(proot,k)
        if self.res != None:
            return self.res.val
        else:
            return -1