请问下如何响应鼠标离开事件？ [复制链接]

大家好，我在一个窗体上有个按钮，自己想做到当鼠标进入但不点击的时候显示一种颜色，鼠标离开按钮的时候又显示回原来的颜色，现在能实现当鼠标进入的时候改变颜色，可是当鼠标离开的时候改变不了，请问下怎么响应到鼠标离开按钮事件？注意鼠标都没有点击的。谢谢。代码如下  
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                                                                                                                                                                                                                                                                                                                                          LiveVideoButton = new QPushButton(QIcon("images/cut.png"),tr(""),this);
    LiveVideoButton->resize(100,50);
    LiveVideoButton->move(0,0);
    LiveVideoButton->setMouseTracking(true);
    LiveVideoButton->installEventFilter(this);
}

bool main_gui::eventFilter(QObject *target, QEvent *event)
{
    if(event->type() == QEvent::MouseMove)
    {
        if(target == LiveVideoButton)
        {
            LiveVideoButton->setStyleSheet("background-color: #202080;");
        }      
    }
    if(event->type() == QEvent::FocusOut)
    {
        if(target == LiveVideoButton)
        {
            LiveVideoButton->setStyleSheet("background-color: #202020;");
        }
    }
}                    当我鼠标移动到按钮上的时候，能进入事件改变北京颜色，我想鼠标离开按钮的时候又显示另一种颜色，应该怎么修改阿？谢谢了！

void QWidget::leaveEvent ( QEvent * event )   [virtual protected]
This event handler can be reimplemented in a subclass to receive widget leave events which are passed in the event parameter.
A leave event is sent to the widget when the mouse cursor leaves the widget.
See also enterEvent(), mouseMoveEvent(), and event().

请问下具体怎么用的？能给点代码吗?谢谢 我是新手。

可以看看文档，Qt Demo有例子。