数据量大的情况

[BradyXs]

如果数据量上千条的话,计算父子关系很耗时,有什么优化方案吗?

[chendabai]

替换框架内TreeHelper中convetData2Node方法为下方法即可

/**
     * 设置Node间，父子关系;让每两个节点都比较一次，即可设置其中的关系
     */
    private static List<Node> convetData2Node(List<Node> nodes) {
        List<Node> res = convert(nodes);
        return res;
    }

    public static List<Node> convert(List<Node> sourceNodeList) {
        List<Node> destNodeList = new ArrayList<Node>();
        //第一步，找出第一级的节点
        //1.1 统计所有节点的id
        List<String> allIds = new ArrayList<String>();
        Map<String, List<Node>> pIdMap = new HashMap<>();
        for (Node tempNode : sourceNodeList) {
            allIds.add((String) tempNode.getId());
            //构造一个存储Pid得map.递归中直接使用键值查找
            String pid = (String) tempNode.getpId();
            if (pIdMap.containsKey(pid)) {
                pIdMap.get(pid).add(tempNode);
            } else {
                List<Node> valueList = new ArrayList<>();
                valueList.add(tempNode);
                pIdMap.put(pid, valueList);
            }

        }

        //所有父节点找不到对应的都是一级id
        for (Node sourceNode : sourceNodeList) {
            if (!allIds.contains(sourceNode.getpId())) {
                //从每个一级节点，递归查找children
                Node destNode = new Node();
                destNode.setId(sourceNode.getId());
                destNode.setName(sourceNode.getName());
                destNode.setpId(sourceNode.getpId());
                destNode.setLevel(1);
                destNode.bean = sourceNode.bean;
                destNode.iconExpand = sourceNode.iconExpand;
                destNode.iconNoExpand = sourceNode.iconNoExpand;
                List<Node> myChilds = getChilderen(destNode, pIdMap);
                destNode.setChildren(myChilds.isEmpty() ? new ArrayList<>() : myChilds);
                destNodeList.add(destNode);
            }
        }
        return destNodeList;
    }


    //    递归获取子节点
    private static List<Node> getChilderen(Node parentNode, Map<String, List<Node>> pIdMap) {
        List<Node> childrenList = new ArrayList<Node>();
        //使用map来进行查找优化效率
        if (!pIdMap.containsKey(parentNode.getId())) {
            return childrenList;
        } else {
            List<Node> sourceNode = pIdMap.get(parentNode.getId());
            for (Node node : sourceNode) {
                Node children = new Node();
                children.setId(node.getId());
                children.setName(node.getName());
                children.setpId(node.getpId());
                children.setLevel(node.getLevel() + 1);
                children.bean = node.bean;
                children.iconExpand = node.iconExpand;
                children.iconNoExpand = node.iconNoExpand;
                children.setParent(parentNode);
                List<Node> myChilds = getChilderen(children, pIdMap);
                children.setChildren(myChilds.isEmpty() ? new ArrayList<>() : myChilds);
                childrenList.add(children);
            }

            return childrenList;
        }

    }