• 最近有个需求，懒加载树形结构，默认展开二级节点。试了获取节点设置节点的expanded属性，发现不生效this.$refs.tree.store.nodesMap[data.id].expanded = true
• 看了下官方文档利用default-expanded-keys属性，懒加载第一级的时候将第一级节点的key放到定义的默认展开数组即default-expanded-keys动态绑定的数组就可以实现改工能
• 部分代码入下图

//css
<el-tree
      ref="menuTree"
       :props="defaultProps"
       node-key="id"
       :key="treeKey"
       lazy
       :default-expanded-keys="treeExpandelKeys"
       :load="loadDeptNode"
       :highlight-current="true"
       @node-contextmenu="rightClick"
       @node-expand="nodeExpand"
        ></el-tree>
//js
loadDeptNode(node, resolve, cb) {
      this.resolve = resolve;
      // 查询参数
      let reqData = {
        roleTypeId: this.roleTypeId,
      };
      if (node.level === 0) {
        this.rootNode = node;
        this.rootNode_resolve = resolve; //同上，把node.level == 0的resolve也存起来
        resolve([this.rootNodeData]);
        this.treeExpandelKeys.push(this.rootNodeData.id)
        this.fristNodeKeys.push(this.rootNodeData.id)
      } else if (node.level > 0) {
        // 向后台发送查询请求
        getNodesService({ params: { node: node.data.id } })
          .then((res) => {
            // 请求成功,且查询出有数据时
            if (res && res.data) {
              let resData = res.data || [];
              var newN = [];
              resData.forEach((it) => {
                let newIt = Object.assign({}, it, {
                  leaf: it.leaf == "1" ? true : false,
                });
                newN.push(newIt);
              });
              if(node.level == 1){
                this.secondNodeKeys = [];
                newN.forEach(item=>{
                    this.secondNodeKeys.push(item.id)
                })
              }
              resolve(newN);
            }
            // 请求失败的情况,或者是查无数据的时候
            else {
              resolve([]);
            }
          })
          .catch((e) => {
            if (node.level === 0) {
              resolve(treeData);
            } else {
              resolve([]);
            }
          });
      }
    },