数据结构遍历的意义
树的遍历
![](https://img.haomeiwen.com/i13248401/0387df3e33d3d032.png)
图的遍历
![](https://img.haomeiwen.com/i13248401/d91781a73d96dd95.png)
树的前序遍历
![](https://img.haomeiwen.com/i13248401/42a2f58287c9c365.png)
![](https://img.haomeiwen.com/i13248401/5e6e1aee28bb726f.png)
图遍历和树遍历区别
![](https://img.haomeiwen.com/i13248401/447e2f8c635ff3db.png)
知识回顾
![](https://img.haomeiwen.com/i13248401/6ad5788e2d8d391d.png)
![](https://img.haomeiwen.com/i13248401/04a2b9a7b63767d3.png)
树的深度优先遍历
![](https://img.haomeiwen.com/i13248401/3e7d5dd109e02359.png)
![](https://img.haomeiwen.com/i13248401/2c04a42f883af1c2.png)
![](https://img.haomeiwen.com/i13248401/00ec6c213332bc9e.png)
普通函数和递归函数调用的区别
![](https://img.haomeiwen.com/i13248401/53d51e43e31351c2.png)
图的深度优先遍历的过程
![](https://img.haomeiwen.com/i13248401/58ec306984f52fa3.png)
初始visited数组都是fasle,当从顶点0开始调用时, 如果没有被遍历,那么遍历结果列表中添加这个顶点,然后开始for循环,执行dfs(1)
![](https://img.haomeiwen.com/i13248401/e8c165616ba14490.png)
Java实现图的DFS
g.txt
7 8
0 1
0 2
1 3
1 4
2 3
2 6
3 5
5 6
import java.io.File;
import java.io.IOException;
import java.util.TreeSet;
import java.util.Scanner;
/// 暂时只支持无向无权图
public class Graph {
private int V;
private int E;
private TreeSet<Integer>[] adj;
public Graph(String pathStr){
File file = new File(pathStr);
try(Scanner scanner = new Scanner(file)){
V = scanner.nextInt();
if(V < 0) throw new IllegalArgumentException("V must be non-negative");
adj = new TreeSet[V];
for(int i = 0; i < V; i ++)
adj[i] = new TreeSet<Integer>();
E = scanner.nextInt();
if(E < 0) throw new IllegalArgumentException("E must be non-negative");
for(int i = 0; i < E; i ++){
int a = scanner.nextInt();
validateVertex(a);
int b = scanner.nextInt();
validateVertex(b);
if(a == b) throw new IllegalArgumentException("Self Loop is Detected!");
if(adj[a].contains(b)) throw new IllegalArgumentException("Parallel Edges are Detected!");
adj[a].add(b);
adj[b].add(a);
}
}
catch(IOException e){
e.printStackTrace();
}
}
private void validateVertex(int v){
if(v < 0 || v >= V)
throw new IllegalArgumentException("vertex " + v + "is invalid");
}
public int V(){
return V;
}
public int E(){
return E;
}
public boolean hasEdge(int v, int w){
validateVertex(v);
validateVertex(w);
return adj[v].contains(w);
}
public Iterable<Integer> adj(int v){
validateVertex(v);
return adj[v];
}
public int degree(int v){
validateVertex(v);
return adj[v].size();
}
@Override
public String toString(){
StringBuilder sb = new StringBuilder();
sb.append(String.format("V = %d, E = %d\n", V, E));
for(int v = 0; v < V; v ++){
sb.append(String.format("%d : ", v));
for(int w : adj[v])
sb.append(String.format("%d ", w));
sb.append('\n');
}
return sb.toString();
}
public static void main(String[] args){
Graph g = new Graph("g.txt");
System.out.print(g);
}
}
GraphDFS.java
package com.duanxuehan.dfs05;
import java.util.ArrayList;
import java.util.Arrays;
/**
* @author Eric Lee
* @date 2020/11/5 20:52
*/
public class GrahDFS {
// 基本图表示
private Graph G;
// 该节点是否被遍历
private boolean[] visited;
// 每个顶点被遍历到的顺序
private ArrayList<Integer> order = new ArrayList<>();
// 定义构造函数,直接在构造函数中执行 dfs
public GrahDFS(Graph G){
// 初始化 图
this.G = G;
// 初始化 是否被遍历过的数组,长度和顶点个数一样
// 默认false也就是没有被遍历过
visited = new boolean[G.V()];
System.out.println(Arrays.toString(visited));
// 直接从顶点0 开始进 dfs
dfs(0);
}
public void dfs(int v){
System.out.println("开始深度优先遍历"+v+"顶点");
// 能进入这里说明v顶点接下来要被遍历了
visited[v] = true;
// 向记录顶点遍历顺序的 order列表中添加此顶点
order.add(v);
// 遍历这个顶点所有的邻接顶点
for(int w : G.adj(v)){
// 每次w代表 和v相邻顶点中的一个
// 如果这个顶点没有被遍历过, 继续调用该顶点的深度优先
if (!visited[w]){
dfs(w);
}
}
}
public ArrayList<Integer> order(){
return order;
}
public static void main(String[] args) {
Graph g = new Graph("g.txt");
GrahDFS grahDFS = new GrahDFS(g);
System.out.println(grahDFS.order());
}
}
DFS的改进
如果是如下的图结构
![](https://img.haomeiwen.com/i13248401/9dd7623f1ff18a6d.png)
对应的txt
7 6
0 1
0 2
1 3
1 4
2 3
2 6
![](https://img.haomeiwen.com/i13248401/1cc4aeccfcb2b8f5.png)
执行dfs之后我们发现 顶点5并没有被遍历,这是因为我们dfs是从0开始的
只调用一次, 5和其他顶点并没有联通, 所以应该执行dfs(5)
![](https://img.haomeiwen.com/i13248401/30be2a3ae2dbf2b4.png)
![](https://img.haomeiwen.com/i13248401/6eb5949a1b09e07e.png)
求联通分量的个数
CC.java
package com.duanxuehan.liantongfenliang06;
import com.duanxuehan.dfs05.Graph;
import java.util.Arrays;
/**
* 求解联通分量的个数
*/
public class CC {
// 定义表示
private Graph G;
// 定义是否被遍历数组
private int[] visited;
// 定义一个计数变量,统计联通分量的个数
private int cccount = 0;
// 构造函数
public CC(Graph G){
this.G = G;
// 初始化布尔数组的值
visited = new int[G.V()];
// 默认将数组中所有值变成 -1, 代表没有被遍历过
for (int i = 0; i < visited.length; i++) {
visited[i] = -1;
}
System.out.println("visited初始值");
System.out.println(Arrays.toString(visited));
// 循环调用 每个顶点的dfs
for (int v = 0; v < G.V() ; v++) {
if (visited[v] == -1){
dfs(v, cccount);
cccount ++;
}
}
}
// 深度优先过程
public void dfs(int v, int ccid){
// 使用ccid表示是否被遍历过,
// 只要不是-1证明已经被遍历过,
// 具体ccid的值实际代表的是所属第几个联通分量
visited[v] = ccid;
// 遍历顶点v所有相邻顶点
for (int w :G.adj(v)){
// visited[w]是 -1 代表w顶点没有被遍历过,要继续进行dfs
if (visited[w] == -1){
dfs(w,ccid);
}
}
}
// 编写专门 统计联通分量个数的方法
public int count(){
for (int e: visited){
System.out.print(e + " ");
}
System.out.println();
return cccount;
}
public static void main(String[] args) {
Graph graph = new Graph("g2.txt");
CC cc = new CC(graph);
System.out.println(Arrays.toString(cc.visited));
System.out.println(cc.count());
}
}
![](https://img.haomeiwen.com/i13248401/1d961e0eaa17ccd9.png)
g2.txt
9 7
0 1
0 2
1 3
1 4
2 3
2 6
7 8
联通分量的改进,列出所有的联通分量详情
CC2.java
package com.duanxuehan.liantongfenliang06;
import com.duanxuehan.dfs05.Graph;
import java.lang.reflect.Array;
import java.util.ArrayList;
// 联通分量的改进,列出所有的联通分量详情
public class CC2 {
// 定义图表示
private Graph G;
// 定义是否被遍历数组
private int[] visited;
// 定义一个计数变量,统计联通分量的个数
private int cccount = 0;
// 构造函数
public CC2(Graph G){
this.G = G;
visited = new int[G.V()];
for (int i = 0; i < visited.length; i++) {
visited[i] = -1;
}
for (int v = 0; v < G.V(); v++) {
if (visited[v] == -1){
dfs(v, cccount);
cccount++;
}
}
}
private void dfs(int v, int ccid){
visited[v] = ccid;
for (int w : G.adj(v)){
if (visited[w] == -1){
dfs(w, ccid);
}
}
}
public int count(){
return cccount;
}
// 判断 顶点v 和顶点 t是否联通
public boolean isConnected(int v, int t){
G.validateVertex(v);
G.validateVertex(t);
return visited[v] == visited[t];
}
// 返回 所有联通分量详情
public ArrayList<Integer>[] compnents(){
// 将所有的联通分量存储成一个数组
// 每一个联通分量的值一个ArrayList
// 一个ArrayList 存储该联通分量的所有值
ArrayList<Integer>[] res = new ArrayList[cccount];
// 为res初始化
for (int i = 0; i < cccount; i++) {
res[i] = new ArrayList<Integer>();
}
// 往res里面进行值添加, 添加具体的联通分量
for (int v = 0; v < G.V() ; v++) {
// visited[v] 代表第几个联通分量
// 向列表中添加元素
res[visited[v]].add(v);
}
return res;
}
public static void main(String[] args) {
Graph graph = new Graph("g2.txt");
CC2 cc2 = new CC2(graph);
System.out.println(cc2.cccount);
System.out.println(cc2.isConnected(7, 8));;
System.out.println(cc2.isConnected(0, 5));
// 显示所以联通分量
ArrayList<Integer>[] compnents = cc2.compnents();
for (int i = 0; i < compnents.length; i++) {
System.out.print("第"+i+"个联通分量" +" : ");
for(int w: compnents[i]){
System.out.print(w + " ");
}
System.out.println();
}
}
}
![](https://img.haomeiwen.com/i13248401/982e206aaa006969.png)
![](https://img.haomeiwen.com/i13248401/a090ad9b57b12e6d.png)
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