题解 | #二叉搜索树的第k个节点#

由于是二插搜索树，进行中序遍历得到有序数组，再对特殊值进行判断

# 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:
    def __init__(self):
        self.nodes = []
        
    def KthNode(self , proot: TreeNode, k: int) -> int:
        # write code here
        if not proot or k == 0:
            return -1
        self.inOrder(proot)
        if k > len(self.nodes):
            return -1
        return self.nodes[k - 1]
        
    def inOrder(self, root):
        if not root:
            return -1
        self.inOrder(root.left)
        self.nodes.append(root.val)
        self.inOrder(root.right)