以下是一个示例代码，演示如何遍历记录以建立层级结构：

class Node:
    def __init__(self, name):
        self.name = name
        self.children = []

def build_hierarchy(records):
    # 创建根节点
    root = Node("root")
    # 创建记录字典，用于快速查找节点
    record_dict = {"root": root}

    for record in records:
        parent_name, child_name = record["parent"], record["name"]
        parent_node = record_dict.get(parent_name)
        if not parent_node:
            # 如果父节点不存在，创建父节点
            parent_node = Node(parent_name)
            record_dict[parent_name] = parent_node
        # 创建子节点
        child_node = Node(child_name)
        # 将子节点添加到父节点的children列表中
        parent_node.children.append(child_node)
        # 将子节点添加到记录字典中
        record_dict[child_name] = child_node

    return root

# 示例输入记录
records = [
    {"name": "A", "parent": "root"},
    {"name": "B", "parent": "A"},
    {"name": "C", "parent": "A"},
    {"name": "D", "parent": "B"},
    {"name": "E", "parent": "B"},
    {"name": "F", "parent": "C"},
    {"name": "G", "parent": "C"},
]

# 构建层级结构
root = build_hierarchy(records)

# 打印层级结构
def print_hierarchy(node, level=0):
    print("  " * level + node.name)
    for child in node.children:
        print_hierarchy(child, level + 1)

print_hierarchy(root)

运行以上代码，将输出以下层级结构：

root
  A
    B
      D
      E
    C
      F
      G