Java 8 中的 Stream 輕松遍歷樹形結(jié)構(gòu)，是真的牛逼！

可能平常會遇到一些需求，比如構(gòu)建菜單，構(gòu)建樹形結(jié)構(gòu)，數(shù)據(jù)庫一般就使用父id來表示，為了降低數(shù)據(jù)庫的查詢壓力，我們可以使用Java8中的Stream流一次性把數(shù)據(jù)查出來，然后通過流式處理。

我們一起來看看，代碼實現(xiàn)為了實現(xiàn)簡單，就模擬查看數(shù)據(jù)庫所有數(shù)據(jù)到List里面。

實體類：Menu.java

/**
?*?Menu
?*
?*?@author?lcry
?*?@date?2020/06/01?20:36
?*/
????@Data
????@Builder
????public?class?Menu?{
????/**
?????*?id
?????*/
?????public?Integer?id;
?????/**
?????*?名稱
?????*/
?????public?String?name;
?????/**
?????*?父id?，根節(jié)點為0
?????*/
?????public?Integer?parentId;
?????/**
?????*?子節(jié)點信息
?????*/
?????public?List?childList;


????public?Menu(Integer?id,?String?name,?Integer?parentId)?{
????????this.id?=?id;
????????this.name?=?name;
????????this.parentId?=?parentId;
????}
????
????public?Menu(Integer?id,?String?name,?Integer?parentId,?List?childList)?{
????????this.id?=?id;
????????this.name?=?name;
????????this.parentId?=?parentId;
????????this.childList?=?childList;
????}

}

遞歸組裝樹形結(jié)構(gòu)：

@Test
public?void?testtree(){
????//模擬從數(shù)據(jù)庫查詢出來
????List?menus?=?Arrays.asList(
????????????new?Menu(1,"根節(jié)點",0),
????????????new?Menu(2,"子節(jié)點1",1),
????????????new?Menu(3,"子節(jié)點1.1",2),
????????????new?Menu(4,"子節(jié)點1.2",2),
????????????new?Menu(5,"根節(jié)點1.3",2),
????????????new?Menu(6,"根節(jié)點2",1),
????????????new?Menu(7,"根節(jié)點2.1",6),
????????????new?Menu(8,"根節(jié)點2.2",6),
????????????new?Menu(9,"根節(jié)點2.2.1",7),
????????????new?Menu(10,"根節(jié)點2.2.2",7),
????????????new?Menu(11,"根節(jié)點3",1),
????????????new?Menu(12,"根節(jié)點3.1",11)
????);

????//獲取父節(jié)點
????List?collect?=?menus.stream().filter(m?->?m.getParentId()?==?0).map(
????????????(m)?->?{
????????????????m.setChildList(getChildrens(m,?menus));
????????????????return?m;
????????????}
????).collect(Collectors.toList());
????System.out.println("-------轉(zhuǎn)json輸出結(jié)果-------");
????System.out.println(JSON.toJSON(collect));
}

/**
?*?遞歸查詢子節(jié)點
?*?@param?root??根節(jié)點
?*?@param?all???所有節(jié)點
?*?@return?根節(jié)點信息
?*/
private?List?getChildrens(Menu?root,?List?all)?{
????List?children?=?all.stream().filter(m?->?{
????????return?Objects.equals(m.getParentId(),?root.getId());
????}).map(
????????????(m)?->?{
????????????????m.setChildList(getChildrens(m,?all));
????????????????return?m;
????????????}
????).collect(Collectors.toList());
????return?children;
}

格式化打印結(jié)果：

原文鏈接：https://blog.csdn.net/qq_19244927/article/details/106481777/

版權(quán)聲明：本文為CSDN博主「Lcry」的原創(chuàng)文章，遵循CC 4.0 BY-SA版權(quán)協(xié)議，轉(zhuǎn)載請附上原文出處鏈接及本聲明。

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