在React中如何在没有事件处理程序的情况下调用函数

Question

如果要满足条件，我想更改状态元素。为此，我尝试编写了一个函数，在该函数中有此条件，在满足条件时有新状态。但是如何调用此函数？我不需要像onClick或onChange这样的事件处理程序。但是怎么可能呢？

this.state = {
    isPresidentVisible: false,
    result: 0,
    minutes: 3,
    seconds: 0,
    isQuizRunning: false,
    isInputDisabled: true,
    isGiveUpVisible: false,
  }
}

  handleChange = (e) => {
    const index = presidents.findIndex(president => (
      president.lastName.toLowerCase() === e.target.value.toLowerCase() || `${president.name.toLowerCase()} ${president.lastName.toLowerCase()}` === e.target.value.toLowerCase()));
    
    if(index !== -1){
      presidents[index].isPresidentVisible = true;
      const isPresidentVisible = presidents[index].isPresidentVisible;
      this.setState(({result}) => ({
        isPresidentVisible,
        result: result + 1
      }))
      e.target.value = '';
    }
  }


  handleClick = () => {
    
    this.setState({
      isQuizRunning: true,
      isInputDisabled: false,
      isGiveUpVisible: true
    })
    this.interval = setInterval(() => {
      const {seconds, minutes} = this.state
      if(seconds > 0){
      this.setState(({seconds}) => ({
        seconds: seconds - 1
      }))
      }
      if(seconds === 0){
        if(minutes === 0){
          this.handleEndGame();
        } else{
          this.setState(({minutes}) => ({
            minutes: minutes - 1,
            seconds: 59
          }))
        }
      }
    },1000)
  }

  handleEndGame = () => {
    clearInterval(this.interval)
    const isPresidentVisible = presidents.map(president => (president.isPresidentVisible = true))
      this.setState({
        isPresidentVisible,
        isInputDisabled: true,
        isGiveUpVisible: false,
    })
  }

我想提出条件：

   checkCondition = () => {
    const {result} = this.state;
    if(result === presidents.length){
      this.handleEndGame();
    }
  }