剑指offer62-二叉搜索树的第k个结点

题目描述

给定一棵二叉搜索树，请找出其中的第k小的结点。

示例1：

输入：   

{5,3,7,2,4,6,8},3

返回值：{4}

说明：    按结点数值大小顺序第三小结点的值为4 

知识点

1）二叉搜索树的中序遍历，找第k个

代码

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回对应节点TreeNode
    def KthNode(self, pRoot, k):
        # write code here重点是写出中序遍历，左子树-根节点-右子树
        a=[]
        a=self.in_search(pRoot,a)
        if k<=0 or k>len(a):
            return None
        return a[k-1]

    def in_search(self,tree,a):
        if tree:
            self.in_search(tree.left,a)
            a.append(tree)
            self.in_search(tree.right,a)
        return a

这里以下写法会报错：'NoneType' object has no attribute 'append'，不知道什么原因

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回对应节点TreeNode
    def KthNode(self, pRoot, k):
        # write code here重点是写出中序遍历，左子树-根节点-右子树
        a=[]
        a=self.in_search(pRoot,a)
        if k<=0 or k>len(a):
            return None
        return a[k-1]

    def in_search(self,tree,a):
        if tree.left:
            self.in_search(tree.left,a)
        a.append(tree)
        if tree.right:
            self.in_search(tree.right,a)
        return a