我試圖通過一些簡單的圖表更好地理解 golang。我在課堂上嘗試過這個，但我無法真正弄清楚。問題是我知道解決方案非常簡單。我試圖讓它通過讀取邊緣來打印每條單獨的路徑。我已經玩過相當多的代碼了，這確實是我能得到的最簡單的形式。package mainimport "fmt"type Graph struct {    Edges map[string][]string}// NewGraph: Create graph with n nodes.func NewGraph() *Graph {    return &Graph{        Edges: make(map[string][]string),    }}// AddEdge: Add an edge from u to v.func (g *Graph) AddEdge(u, v string) {    g.Edges[u] = append(g.Edges[u], v)}func (g *Graph) Walk(u string) {    for _, v := range g.Edges[u] {            fmt.Printf("%s -> %s", u, v)        g.Walk(v)    }}func (g *Graph) adjacentEdgesExample() {    fmt.Println("Printing all edges in graph.")    for u, adjacent := range g.Edges { // Nodes are labelled 0 to N-1.        for _, v := range adjacent {            // Edge exists from u to v.            fmt.Printf("Edge: %s -> %s\n", u, v)        }    }}func main() {    g := NewGraph()    g.AddEdge("A", "B")    g.AddEdge("A", "D")    g.AddEdge("D", "E")    g.AddEdge("B", "C")    g.adjacentEdgesExample()    fmt.Println(g.Edges)    g.Walk("A")}示例位于：https://goplay.space/#Ro1puZYgu5X其結果是：Printing all edges in graph.Edge: B -> CEdge: A -> BEdge: A -> DEdge: D -> Emap[A:[B D] B:[C] D:[E]]A -> BB -> CA -> DD -> E% 我希望看到 Walk 方法做這樣的事情：A -> B -> CA -> D -> E