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reactjs - Is it possible to return empty in react render function?

I have a notification component, and I have a timeout setting for it. After timeout I call this.setState({isTimeout:true}).

What I want to do is if already timeout, I want just render nothing:

render() {
  let finalClasses = "" + (this.state.classes || "");
  if (isTimeout){
    return (); // here has some syntax error
  }
  return (<div>{this.props.children}</div>);
}

The problem is:

return (); // here has some syntax error

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Yes you can, but instead of blank, simply return null if you don't want to render anything from component, like this:

return (null);

Another important point is, inside JSX if you are rendering element conditionally, then in case of condition=false, you can return any of these values false, null, undefined, true. As per DOC:

booleans (true/false), null, and undefined are valid children, they will be Ignored means they simply don’t render.

All these JSX expressions will render to the same thing:

<div />

<div></div>

<div>{false}</div>

<div>{null}</div>

<div>{undefined}</div>

<div>{true}</div>

Example:

Only odd values will get rendered, because for even values we are returning null.

const App = ({ number }) => {
  if(number%2) {
    return (
      <div>
        Number: {number}
      </div>
    )
  }
  
  return (null);           //===> notice here, returning null for even values
}

const data = [1,2,3,4,5,6];

ReactDOM.render(
  <div>
    {data.map(el => <App key={el} number={el} />)}
  </div>,
  document.getElementById('app')
)
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react-dom.min.js"></script>

<div id='app' />

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