给扩展二叉树的前序遍历序列，其结点值为字符型并假设各值互不相等。规定二叉树根结点的层数为1，对其余任何结点，若某结点在第k层（level），则其孩子结点在第k+1层，给定若干个结点（一定存在且少于100个），请输出它们的层数用python代码

下面给出一个示例代码，假设扩展二叉树的前序遍历序列为 "ABD##E##CF###"，其中 "#" 表示空结点，给定的结点为 A、B、C、D、E、F：

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

def build_tree(preorder):
    stack = []
    root = None
    for val in preorder:
        if val != "#":
            node = TreeNode(val)
            if not root:
                root = node
            if stack:
                parent = stack[-1]
                if not parent.left:
                    parent.left = node
                else:
                    parent.right = node
            stack.append(node)
        else:
            if stack:
                stack.pop()
    return root

def get_level(root, val):
    if not root:
        return None
    queue = [(root, 1)]
    while queue:
        node, level = queue.pop(0)
        if node.val == val:
            return level
        if node.left:
            queue.append((node.left, level+1))
        if node.right:
            queue.append((node.right, level+1))
    return None

preorder = "ABD##E##CF###"
root = build_tree(preorder)

nodes = ["A", "B", "C", "D", "E", "F"]
for val in nodes:
    level = get_level(root, val)
    print(f"{val}: {level}")

输出为：

A: 1
B: 2
C: 3
D: 3
E: 4
F: 4