java实现 tree结构模糊搜索功能

 //原始数据先转成树结构

    private List<Organization> recursionMethodOrg(List<Organization> treeList) {
        List<Organization> trees = new ArrayList<>();
        for (Organization tree : treeList) {
            // 找出父节点
            if ("0".equals(tree.getOrgParentId())) {
                // 调用递归方法填充子节点列表
                trees.add(findChildrenOrg(tree, treeList));
            }
        }
        return trees;
    }

    private Organization findChildrenOrg(Organization tree, List<Organization> treeList) {
        for (Organization node : treeList) {
            if (tree.getOrgId().equals(node.getOrgParentId())) {
                if (tree.getChildren() == null) {
                    tree.setChildren(new ArrayList<>());
                }
                // 递归 调用自身
                tree.getChildren().add(findChildrenOrg(node, treeList));
            }
        }
        return tree;
    }


    //再通过树结构去模糊查询
    private List<Organization> filterTreeByKeyWordByOrg(List<Organization> data, String keyword) {

        if (CollectionUtils.isEmpty(data)) {
            return data;
        }
        filterByOrg(data, keyword);
        return data;
    }

    /**
     * 递归方法
     * @param list 任意层级的节点
     * @param keyword 关键字
     */
    private void filterByOrg(List<Organization> list, String keyword){
        Iterator<Organization> parent = list.iterator();
        while (parent.hasNext()){
            //当前节点
            Organization t = parent.next();
            if (StrUtil.isNotEmpty(keyword) && !t.getOrgName().contains(keyword)) {
                //当前节点不包含关键字，继续遍历下一级
                // 取出下一级节点
                List<Organization> children = t.getChildren();
                // 递归
                if (!CollectionUtils.isEmpty(children)) {
                    filterByOrg(children, keyword);
                }
                //下一级节点都被移除了，那么父节点也移除，因为父节点也不包含关键字
              if (CollectionUtils.isEmpty(t.getChildren())) {
                    parent.remove();
                }
            } else {
                //当前节点包含关键字，继续递归遍历
                //子节点递归如果不包含关键字则会进入if分支被删除
                List<Organization> children = t.getChildren();
                // 递归
                if (!CollectionUtils.isEmpty(children)) {
                    filterByOrg(children, keyword);
                }
            }
        }
    }