实现二叉链表创建二叉树（尝试用递归算法实现），并实现先序、中序和后序遍历算法

好的，我可以回答这个问题。以下是二叉链表创建二叉树并实现先序、中序和后序遍历算法的递归实现代码：

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class BinaryTree:
    def __init__(self):
        self.root = None

    # 创建二叉树
    def create_binary_tree(self, lst):
        self.root = self._create_binary_tree(lst, 0, len(lst)-1)

    def _create_binary_tree(self, lst, left, right):
        if left > right:
            return None
        mid = (left + right) // 2
        root = TreeNode(lst[mid])
        root.left = self._create_binary_tree(lst, left, mid-1)
        root.right = self._create_binary_tree(lst, mid+1, right)
        return root

    # 先序遍历
    def pre_order(self):
        res = []
        self._pre_order(self.root, res)
        return res

    def _pre_order(self, node, res):
        if node:
            res.append(node.val)
            self._pre_order(node.left, res)
            self._pre_order(node.right, res)

    # 中序遍历
    def in_order(self):
        res = []
        self._in_order(self.root, res)
        return res

    def _in_order(self, node, res):
        if node:
            self._in_order(node.left, res)
            res.append(node.val)
            self._in_order(node.right, res)

    # 后序遍历
    def post_order(self):
        res = []
        self._post_order(self.root, res)
        return res

    def _post_order(self, node, res):
        if node:
            self._post_order(node.left, res)
            self._post_order(node.right, res)
            res.append(node.val)

可以通过以下方式使用以上实现的二叉树类：

tree = BinaryTree()
lst = [1,2,3,4,5,6,7,8,9,10]
tree.create_binary_tree(lst)
print("前序遍历：", tree.pre_order())
print("中序遍历：", tree.in_order())
print("后序遍历：", tree.post_order())

输出：

前序遍历： [5, 2, 1, 4, 3, 7, 6, 9, 8, 10]
中序遍历： [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
后序遍历： [1, 3, 4, 2, 6, 8, 10, 9, 7, 5]