如何使用onClick函数设置子组件中的父组件的状态？

// 父组件
import React, { useState } from 'react';
import ChildComponent from './ChildComponent';

const ParentComponent = () => {
  const [parentState, setParentState] = useState(false);

  const updateParentState = () => {
    setParentState(!parentState);
  };

  return (
    <div>
      <ChildComponent parentState={parentState} updateParentState={updateParentState} />
    </div>
  );
};

export default ParentComponent;

// 子组件
import React from 'react';

const ChildComponent = ({ parentState, updateParentState }) => {
  const handleClick = () => {
    updateParentState();
  };

  return (
    <div>
      <button onClick={handleClick}>点击更新父组件状态</button>
      <p>父组件状态：{parentState ? '开启' : '关闭'}</p>
    </div>
  );
};

export default ChildComponent;