java 递归转非递归（树形转非树形）

1.情景展示

接着上一篇讲，在上一篇我们已经将数据通过递归转成了树形结构。

如何将树形结构转非树形结构？（仍然按照树形层级关系进行平铺显示）

如何将递归函数转成迭代函数？

查看代码

[{
		name = 根节点,
		childList = [{
				name = 一级节点1,
				childList = [{
						name = 二级节点1.1,
						childList = [],
						id = 3,
						parentId = 2
					}, {
						name = 二级节点1.2,
						childList = [],
						id = 4,
						parentId = 2
					}, {
						name = 二级节点1.3,
						childList = [],
						id = 5,
						parentId = 2
					}
				],
				id = 2,
				parentId = 1
			}, {
				name = 一级节点2,
				childList = [{
						name = 二级节点2.1,
						childList = [{
								name = 三级节点2.1.1,
								childList = [],
								id = 9,
								parentId = 7
							}, {
								name = 三节点2.1.2,
								childList = [],
								id = 10,
								parentId = 7
							}
						],
						id = 7,
						parentId = 6
					}, {
						name = 二级节点2.2,
						childList = [],
						id = 8,
						parentId = 6
					}
				],
				id = 6,
				parentId = 1
			}, {
				name = 一级节点3,
				childList = [{
						name = 二级节点3.1,
						childList = [{
								name = 三级节点3.1.1,
								childList = [{
										name = 四级节点3.1.1.1,
										childList = [{
												name = 五级节点3.1.1.1.1,
												childList = [],
												id = 15,
												parentId = 14
											}
										],
										id = 14,
										parentId = 13
									}
								],
								id = 13,
								parentId = 12
							}
						],
						id = 12,
						parentId = 11
					}
				],
				id = 11,
				parentId = 1
			}
		],
		id = 1,
		parentId = 0
	}
]

2.具体分析

可以通过栈Stack来完成。

3.解决方案

代码实现

/*
 * 将父节点下的子节点进行由嵌套改成平铺
 * @description: 递归转迭代
 * 按照层级关系，将嵌套改成平铺
 * @param: root 嵌套数据
 * @return: java.util.ArrayList<java.util.Map<java.lang.String,java.lang.Object>>
 * 平铺数据
 */
public static ArrayList<Map<String, Object>> tileTraversal(Map<String, Object> root) {
    ArrayList<Map<String, Object>> res = new ArrayList<>();
    if (root == null) {
        return res;
    }

    // 建立一个新的栈实例
    Stack<Map<String, Object>> stack = new Stack<>();
    // 把原始内容压入新建栈
    stack.push(root);
    while (!stack.empty()) {
        // 出栈，将栈顶元素赋给temp
        Map<String, Object> temp = stack.pop();
        Map<String, Object> copyMap = new HashMap<>(temp.size());
        // 复制当前节点
        copyMap.putAll(temp);
        // 移除map（该节点）当中的子节点
        copyMap.remove("childList");
        // 将此节点放到集合当中
        res.add(copyMap);

        // 当前节点，存在子节点
        if (temp.get("childList") != null && ((List<Map<String, Object>>) temp.get("childList")).size() > 0) {
            // ((List<Map<String, Object>>) temp.get("childList")).forEach(m -> stack.push(m));

            List<Map<String, Object>> list = (List<Map<String, Object>>) temp.get("childList");
            for (int i = list.size() - 1; i >= 0; i--) {
                // 将子节点倒序进栈
                stack.push(list.get(i));
            }
        }
    }
    return res;
}

调用

// mapList是树形结构（已经存在的数据）
List<Map<String, Object>> tileList = new ArrayList<>();
mapList.forEach(m -> tileList.addAll(tileTraversal(m)));
System.out.println(tileList);

[{name=根节点, id=1, parentId=0}, {name=一级节点1, id=2, parentId=1}, {name=二级节点1.1, id=3, parentId=2}, {name=二级节点1.2, id=4, parentId=2}, {name=二级节点1.3, id=5, parentId=2}, {name=一级节点2, id=6, parentId=1}, {name=二级节点2.1, id=7, parentId=6}, {name=三级节点2.1.1, id=9, parentId=7}, {name=三节点2.1.2, id=10, parentId=7}, {name=二级节点2.2, id=8, parentId=6}, {name=一级节点3, id=11, parentId=1}, {name=二级节点3.1, id=12, parentId=11}, {name=三级节点3.1.1, id=13, parentId=12}, {name=四级节点3.1.1.1, id=14, parentId=13}, {name=五级节点3.1.1.1.1, id=15, parentId=14}]

4.效果展示

转JSON

System.out.println(JSON.toJSON(tileList));

[{"name":"根节点","id":"1","parentId":"0"},{"name":"一级节点1","id":"2","parentId":"1"},{"name":"二级节点1.1","id":"3","parentId":"2"},{"name":"二级节点1.2","id":"4","parentId":"2"},{"name":"二级节点1.3","id":"5","parentId":"2"},{"name":"一级节点2","id":"6","parentId":"1"},{"name":"二级节点2.1","id":"7","parentId":"6"},{"name":"三级节点2.1.1","id":"9","parentId":"7"},{"name":"三节点2.1.2","id":"10","parentId":"7"},{"name":"二级节点2.2","id":"8","parentId":"6"},{"name":"一级节点3","id":"11","parentId":"1"},{"name":"二级节点3.1","id":"12","parentId":"11"},{"name":"三级节点3.1.1","id":"13","parentId":"12"},{"name":"四级节点3.1.1.1","id":"14","parentId":"13"},{"name":"五级节点3.1.1.1.1","id":"15","parentId":"14"}]

格式化

查看代码


[{
	"name": "根节点",
	"id": "1",
	"parentId": "0"
},
{
	"name": "一级节点1",
	"id": "2",
	"parentId": "1"
},
{
	"name": "二级节点1.1",
	"id": "3",
	"parentId": "2"
},
{
	"name": "二级节点1.2",
	"id": "4",
	"parentId": "2"
},
{
	"name": "二级节点1.3",
	"id": "5",
	"parentId": "2"
},
{
	"name": "一级节点2",
	"id": "6",
	"parentId": "1"
},
{
	"name": "二级节点2.1",
	"id": "7",
	"parentId": "6"
},
{
	"name": "三级节点2.1.1",
	"id": "9",
	"parentId": "7"
},
{
	"name": "三节点2.1.2",
	"id": "10",
	"parentId": "7"
},
{
	"name": "二级节点2.2",
	"id": "8",
	"parentId": "6"
},
{
	"name": "一级节点3",
	"id": "11",
	"parentId": "1"
},
{
	"name": "二级节点3.1",
	"id": "12",
	"parentId": "11"
},
{
	"name": "三级节点3.1.1",
	"id": "13",
	"parentId": "12"
},
{
	"name": "四级节点3.1.1.1",
	"id": "14",
	"parentId": "13"
},
{
	"name": "五级节点3.1.1.1.1",
	"id": "15",
	"parentId": "14"
}]

说明：

事实上，使用SQL进行递归查询，返回的就是这种格式的数据，即：按照层级关系排好序展示出来；

如果前端需要的是这种排好序的数据，直接返回即可；

如果需要返回树形结构的话，见文末推荐。

写在最后