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
#
array = []
class Solution:
    def Append(self, t):
        array.append(t.val)
        if t.left != None:
            Solution.Append(self, t.left)
        if t.right != None:
            Solution.Append(self, t.right)
    def KthNode(self , proot: TreeNode, k: int) -> int:
        # write code here
        if proot == None or k == 0:
            return -1
        array.append(proot.val)
        if proot.left != None:
            Solution.Append(self, proot.left)
        if proot.right != None:
            Solution.Append(self, proot.right)
        array.sort()
        if k > len(array):
            return -1
        return array[k - 1]