采用递归的方法实现：
原理：
树函数(列表，父ID)
ret = []
for 节点 in 列表:
if 节点的parent_id 等于 父ID
节点.children = 获取树(列表, 节点ID)
ret = append(ret,节点)
return ret
实例测试：
package main
import (
"encoding/json"
"fmt"
)
type Node struct {
Id int `json:"id"`
ParentId int `json:"parent_id"`
Name string `json:"name"`
Children []*Node `json:"children"`
}
func getTreeRecursive(list []*Node, parentId int) []*Node {
res := make([]*Node, 0)
for _, v := range list {
if v.ParentId == parentId {
v.Children = getTreeRecursive(list, v.Id)
res = append(res, v)
}
}
return res
}
func main() {
list := []*Node{
{4, 3, "ABA", nil},
{3, 1, "AB", nil},
{1, 0, "A", nil},
{2, 1, "AA", nil},
}
res := getTreeRecursive(list, 0)
bytes, _ := json.MarshalIndent(res, "", " ")
fmt.Printf("%s\n", bytes)
}采用循环的方法实现：
原理：
获取树(列表，父ID)
memo = {}
for 节点 in 列表:
//构造memo给节点的父ID查找追加节点用
if 节点ID in memo:
节点.children = memo[节点ID].children //之前构造的children数组覆盖当前节点的children
memo[节点ID] = 节点
else
节点.children = []
memo[节点ID] = 节点
//给像父对象的children追加
if 节点父ID in memo:
memo[节点父ID].children.add(memo[节点ID]) //追加当前构造的ID节点
else:
memo[节点父ID] = {'children':[memo[节点ID]]} //初始化父对象再追加
return memo[父ID].children实现实例：
package main
import (
"encoding/json"
"fmt"
)
type Node struct {
Id int `json:"id"`
ParentId int `json:"parent_id"`
Name string `json:"name"`
Children []*Node `json:"children"`
}
func getTreeIterative(list []*Node, parentId int) []*Node {
memo := make(map[int]*Node)
for _, v := range list {
if _, ok := memo[v.Id]; ok {
v.Children = memo[v.Id].Children
memo[v.Id] = v
} else {
v.Children = make([]*Node, 0)
memo[v.Id] = v
}
if _, ok := memo[v.ParentId]; ok {
memo[v.ParentId].Children = append(memo[v.ParentId].Children, memo[v.Id])
} else {
memo[v.ParentId] = &Node{Children: []*Node{memo[v.Id]}}
}
}
return memo[parentId].Children
}
func main() {
list := []*Node{
{4, 3, "ABA", nil},
{3, 1, "AB", nil},
{1, 0, "A", nil},
{2, 1, "AA", nil},
{5, 3, "ABB", nil},
}
res := getTreeIterative(list, 0)
bytes, _ := json.MarshalIndent(res, "", " ")
fmt.Printf("%s\n", bytes)
}好了，本文内容全部结束了对 go语言实现 菜单树形结构的算法实现。希望对您有所帮助。
