Why does the recursion work this way in Go? - go

Question

I'm learning Go to prepare for a coding interview, and there's a leetcode question for tree traversal. It works with recursion in Python, but Golang behaves differently.
I notice the elements in the res slice gets cleared one by one when every call stack is popped.
The original question is
The result should be [1,3,5,6,2,4], and it was returned correctly in Python, but the following Go code returns []
/**
* Definition for a Node.
* type Node struct {
* Val int
* Children []*Node
* }
*/
func preorder(root *Node) []int {
res := []int{}
traverse(root, res)
return res
}
func traverse(root *Node, res []int){
if root == nil{
return
}
res = append(res, root.Val)
for _, n := range root.Children{
traverse(n, res)
}
// the last element is removed from the slice every time when the code execution reaches here
}
------------------------------------- updates --------------------------------------
Thanks for all your answers, now I think I have a better understanding of how slice works in Go
The following code works for me now:
/**
* Definition for a Node.
* type Node struct {
* Val int
* Children []*Node
* }
*/
func preorder(root *Node) []int {
res := []int{}
res = traverse(root, res)
return res
}
func traverse(root *Node, res []int) []int{
if root == nil{
return res
}
res = append(res, root.Val)
for _, n := range root.Children{
res = traverse(n, res)
}
return res
}

Answer 1

you can consider passing the res as a pointer *[]int instead so that the values get updated to the same slice else you are just passing a copy of it.

Answer 2

In your terminating case, you're not returning the accumulator:
if root == nil{
return
}
But...
I would use a closure so as to avoid cloning the slice on every recursive call (Go passes by value):
type Node struct {
Val int
Children []*Node
}
func preorder(root *Node) (res []int) {
var traverse func(*Node)
traverse = func(root *Node) {
if root != nil {
res = append(res, root.Val)
for _, n := range root.Children {
traverse(n)
}
}
}
traverse(root)
return res
}