2015-02-25 08:47
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Golang中LinkedList <T>的等效项是什么

In my use case, I would like to know how the following Java code would be implemented in Go

class TreeNode {
    public int data;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(){}

LinkedList<TreeNode> treeList = new LinkedList<TreeNode>();

I am able to import the container/list package and add an interface. But it is not allowing any generic object. Do I have to implement my own version of list with TreeNode struct?

I just need to know how LinkedList<T> would work in Go.

EDIT 1: To make it clear, I am adding the complete code here. I am trying to find the linked list of all nodes at each depth in a binary tree. I used two packages list and binary tree. You can find the source code for binarytree here and list here. list is same as container/list but I added few extra functions

package main
import (
func main() {

    inArr := []int{4, 5, 7, 8, 9}
    t1 := binarytree.NewMinimalHeightBST(inArr, 0, len(inArr)-1)
    var nodeList []*list.List

    nodeList = getLevelbasedList(t1, 0)

    for _, value := range nodeList {
        fmt.Print("[ ")
        for x := value.Front(); x != nil; x = x.Next() {
            fmt.Print(x.Value.(int), " ")

func getLevelbasedList(root *binarytree.Tree, level int) []*list.List {
    if root == nil {
        return nil
    var nodeList []*list.List
    parents := list.New()
    current := list.New()


    for current.Len() > 0 {
        nodeList = append(nodeList, current)
        parents = current
        current = list.New()

        for x := current.Front(); x != nil; x = x.Next() {
            node := x.Value.(*binarytree.Tree)
            if node.Left != nil {
                current = current.PushFront(node.Left)
            if node.Right != nil {
                current = current.PushFront(node.Right)
        return nodeList

And the error is,

./question4_4b.go:56: cannot use current.PushFront((interface {})(node.Left)) (type *list.Element) as type *list.List in assignment
./question4_4b.go:59: cannot use current.PushFront((interface {})(node.Right)) (type *list.Element) as type *list.List in assignment

EDIT 2: Based on JamesHenstridge's comment I edited from

current = current.PushFront(node.Left)



And the issue resolved. But now I am getting interface conversion error,

[ panic: interface conversion: interface is *binarytree.Tree, not int

goroutine 1 [running]:
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  • dpql57753 2015-02-25 09:04

    Go doesn't support generic types (see FAQ question Why does Go not have generic types?).

    You have to use Type assertions to obtain the typed value you want.

    E.g. create your TreeNode type:

    type TreeNode struct {
        Data  int
        Left  *TreeNode
        Right *TreeNode

    And to iterate over a list containing TreeNode values:

    l := list.New()
    // Populate list
    for e := l.Front(); e != nil; e = e.Next() {
        if tn, ok := e.Value.(TreeNode); ok {
            // do something with tn which is of type TreeNode
        } else {
            // e.Value is not of type TreeNode

    If you assemble the list and you can be sure it only contains values of type TreeNode, you can omit the error check in the type assertion and it becomes like this:

    for e := l.Front(); e != nil; e = e.Next() {
        // if e.Value would not be of type TreeNode, run-time panic would occur
        tn := e.Value.(TreeNode) // tn is of type TreeNode


    The error you're getting:

    cannot use current.PushFront((interface {})(node.Left)) (type *list.Element)
        as type *list.List in assignment

    At line:

    current = current.PushFront(node.Left)

    The current variable is of type list.List, and the method current.PushFront() returns a value of type *list.Element. These are 2 different types, you can't assign a *Element to a variable that has a type of List.

    Edit 2:

    Your 2nd error:

    panic: interface conversion: interface is *binarytree.Tree, not int

    Is caused by the line:

    fmt.Print(x.Value.(int), " ")

    You try to assert that the value x.Value is of type int but it isn't! x.Value is of type *binarytree.Tree so the assertion will obviously fail.

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