tree 根据子元素获取所有父级目录结构

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  <script>
    let rootTree = {
      childList: [
          title: '1-1', id: 1, childList: [
              title: '1-2', id: 2, childList: [
                  title: '1-3', id: 3, childList: [
                      title: '文件', id: 4
          title: '2-1', id: 11, childList: [
              title: '2-2', id: 111, childList: [
                  title: '2-3', id: 311, childList: [
                      title: '文件', id: 422
          title: '3-1', id: 2222, childList: [
              title: '3-2', id: 33333, childList: [
                  title: '3-3', id: 4444, childList: [
                      title: '文件', id: 5555
    let xx = familyTree(rootTree, 422)
    console.log(xx);
    // 根据子元素返回他的所有父类
    // 查找父节点
    function familyTree (rootTree, id) {
      let temp = []
      let forFn = (treeItem, id) => {
        // treeItem 精髓 - 传入者是一个对象.children结构，而且是作为目录节点，因为在子集childList里面找到以后
        // 我们需要把父级的id进行一个存储，同时把父级的id继续递归，父-父级的childList里面找到以后
        // 又会把父-父级的id进行存储，同时继续开启递归，直到顶层为空
        for (let i = 0; i < treeItem.childList.length; i++) {
          let item = treeItem.childList[i]
          if (item.id === id) {
            // 精髓 - 这边是返回父级的id
            treeItem.id && temp.unshift({ title: treeItem.title, id: treeItem.id })
            forFn(rootTree, treeItem.id)
            break
          } else {
            if (item.childList?.length) {
              // 精髓 - 这边通过children判断，但是传入的是item
              forFn(item, id)
      forFn(rootTree, id)
      return temp
     </script>