请写出按先序次序建立一棵二叉树（二叉链表），然后输出先序遍历结果的第k个结点值的代码

好的，我知道了。以下是按先序次序建立二叉树并输出先序遍历结果的第k个结点值的代码：

class TreeNode:
    def __init__(self, val=0):
        self.val = val
        self.left = None
        self.right = None
        
def build_tree(vals):
    if not vals:
        return None
    
    val = vals.pop(0)
    
    if val == None:
        return None
    
    node = TreeNode(val)
    node.left = build_tree(vals)
    node.right = build_tree(vals)
    
    return node

def pre_order_traversal(node):
    if not node:
        return
    
    print(node.val, end=" ")
    pre_order_traversal(node.left)
    pre_order_traversal(node.right)
    
vals = [1, 2, None, 3, None, None, 4, None, 5]

root = build_tree(vals)

pre_order_traversal(root)

k = 3
count = 0

def kth_node(node):
    global count
    
    if not node:
        return None
    
    count += 1
    
    if count == k:
        return node.val
    
    return kth_node(node.left) or kth_node(node.right)

print(kth_node(root))

以上代码按先序次序建立了一棵二叉树，然后输出了先序遍历结果，并计算并输出了先序遍历结果的第k个结点值。