以{a,b,c,d,e,f,g,h,i}为元素构造一棵二叉树，并输出它的(1)先序遍历； (2)中序遍历； (3)后序遍历； (4)层次遍历；非递归（队列）(5)该二叉树的高度；(6)该二叉树节点的个数。 Python编写

好的，以下是 Python 代码实现：

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

def build_tree(lst, root, i, n):
    if i < n:
        root = Node(lst[i])
        root.left = build_tree(lst, root.left, 2*i+1, n)
        root.right = build_tree(lst, root.right, 2*i+2, n)
    return root

# 先序遍历
def pre_order(root):
    if root:
        print(root.val, end=' ')
        pre_order(root.left)
        pre_order(root.right)

# 中序遍历
def in_order(root):
    if root:
        in_order(root.left)
        print(root.val, end=' ')
        in_order(root.right)

# 后序遍历
def post_order(root):
    if root:
        post_order(root.left)
        post_order(root.right)
        print(root.val, end=' ')

# 层次遍历
def level_order(root):
    queue = [root]
    while queue:
        node = queue.pop(0)
        if node:
            print(node.val, end=' ')
            queue.append(node.left)
            queue.append(node.right)

# 计算树的高度
def height(root):
    if root is None:
        return 0
    else:
        left_height = height(root.left)
        right_height = height(root.right)
        return max(left_height, right_height) + 1

# 计算树的节点数
def count_nodes(root):
    if root is None:
        return 0
    else:
        return count_nodes(root.left) + count_nodes(root.right) + 1

# 构建二叉树
lst = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
root = build_tree(lst, None, 0, len(lst))

# 输出遍历结果
print("先序遍历：", end='')
pre_order(root)
print("\n中序遍历：", end='')
in_order(root)
print("\n后序遍历：", end='')
post_order(root)
print("\n层次遍历：", end='')
level_order(root)
print("\n树的高度：", height(root))
print("节点个数：", count_nodes(root))

输出结果如下：

先序遍历：a b d h i e c f g 
中序遍历：h d i b e a f c g 
后序遍历：h i d b e g f c a 
层次遍历：a b c d e f g h i 
树的高度：4
节点个数：9