突然想起一个面试题，用go实现不太好写，明天在想有什么好的方法实现图，暂时就想到这么实现，具体分析看代码注释
package main
import "fmt"
type list struct {
data string
next []*list //代表每个节点能够访问的节点，比如v0的next为v1,v2,v3
}
var m map[string]bool
func DFS(row []*list){
if len(row) == 0 {
return
}
//跟一个节点沿着一条路径走到头，走到头之后再进行下一个节点
for i := range row {
if ok,_ := m[row[i].data];!ok{
fmt.Print(row[i].data, " ")
m[row[i].data] = true
if len(row[i].next) > 0 {
DFS(row[i].next)
}
}
}
}
func BFS(row []*list){
if len(row) == 0 {
return
}
//下一层的点的集合
nextRow := make([]*list, 0)
//遍历当前层所有点，未被访问则输出并将该点能访问的点加到nextRow
for i := range row {
if ok, _ := m[row[i].data];!ok {
fmt.Print(row[i].data," ")
m[row[i].data] = true
if len(row[i].next) != 0 {
nextRow = append(nextRow, row[i].next...)
}
}
}
BFS(nextRow)
}
func main(){
m = make(map[string]bool)
v0 := &list{data : "v0"}
v1 := &list{data : "v1"}
v2 := &list{data : "v2"}
v3 := &list{data : "v3"}
v4 := &list{data : "v4"}
v5 := &list{data : "v5"}
v6 := &list{data : "v6"}
v0.next = append(v0.next, v1,v2,v3)
v2.next = append(v2.next, v4)
v1.next = append(v1.next, v4,v5)
v3.next = append(v3.next, v5)
v4.next = append(v4.next, v6)
v5.next = append(v5.next, v6)
fmt.Print("DFS : ")
DFS([]*list{v0})
m = make(map[string]bool)
fmt.Print("\nBFS : ")
BFS([]*list{v0})
}
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肯定不是最好的实现方法，欢迎一起讨论
————————————————
层序遍历模板,语言为 golang
func bfs(p *TreeNode) []int {
res := make([]int, 0)
if p == nil {
return res
}
queue := []*TreeNode{p}
for len(queue) > 0 {
length := len(queue)
for length > 0 {
length--
if queue[0].Left != nil {
queue = append(queue, queue[0].Left)
}
if queue[0].Right != nil {
queue = append(queue, queue[0].Right)
}
res = append(res, queue[0].Val)
queue = queue[1:]
}
}
return res
}
