在vue中如何实现页面跳转后返回原页面初始位置

//引用vuex 
import Vuex from 'vuex' 
Vue.use(Vuex)

var store = new Vuex.Store({ 
 state: { 
 recruitScrollY:0 
 }, 
 getters: { 
 recruitScrollY:state => state.recruitScrollY 
 }, 
 mutations: { 
 changeRecruitScrollY(state,recruitScrollY) { 
 state.recruitScrollY = recruitScrollY 
 } 
 }, 
 actions: { 
 
 }, 
 modules: {} 
})

beforeRouteLeave(to, from, next) { 
 let position = window.scrollY //记录离开页面的位置 
 if (position == null) position = 0 
 this.$store.commit('changeRecruitScrollY', position) //离开路由时把位置存起来 
 next() 
 }, 
 watch: { 
 '$route' (to, from) { 
 if (to.name === 'NewRecruit') {//跳转的的页面的名称是"NewRecruit",这里就相当于我们listview页面，或者原始页面 
 let recruitScrollY = this.$store.state.recruitScrollY 
 window.scroll(0, recruitScrollY) 
 } 
 } 
 }

<keep-alive v-if="$route.meta.keepAlive"> 
 <router-view></router-view> 
 </keep-alive> 
 <router-view v-if="!$route.meta.keepAlive"></router-view>

Vue.use(Router) 
 
const routerApp = new Router({ 
 routes: [{ 
 { 
 path: '/NewRecruit', 
 name: 'NewRecruit', 
 component: NewRecruit, 
 meta: { 
 keepAlive: true 
 } 
 }, 
 { 
 path: '/NewRecruitDesc/:id', 
 name: 'NewRecruitDesc', 
 component: NewRecruitDesc, 
 meta: { 
 keepAlive: true 
 } 
 }, 
 { 
 path: '/SubmitSucess', 
 name: 'SubmitSucess', 
 component: SubmitSucess, 
 meta: { 
 keepAlive: false 
 } 
 } 
 ] 
}) 
 
export default routerApp