在树中搜索的关键通常是实现一个很好的 Ajacency 函数，它为特定节点提供相邻节点。
对于这棵树，邻接函数会想要找到节点所在的层级，然后将下一层的节点作为相邻节点返回。它看起来像这样：
/**
* This returns the adjacent nodes to an integer node in the tree
* @param node
* @return a set of adjacent nodes, and null otherwise
*/
public Set<Integer> getAdjacentsToNode(Integer node) {
for (int i = 0; i < tree.size(); i++) {
Set<Integer> level = tree.get(i);
if (level.contains(node) && i < tree.size() - 1) {
return tree.get(i + 1);
}
}
return null;
}
假设我们已经将树定义为类中的一个字段。
在我们运行搜索之前，我们希望适当地初始化设置根目录并对路径做一些准备工作。因此，我们执行以下操作：
/**
* initializes our search, sets the root and adds it to the path
*/
public void initialize() {
root = -1;
for (Integer node : tree.get(0)) {
root = node;
}
currentPath.add(root);
}
假设currentPath 和root 已经定义为字段。
然后，我们在树上运行 DFS 搜索，在遍历树时将每个节点附加到当前路径，并将该路径添加到我们设置的路径中，并在到达死角（叶子）时重置它：
/**
* runs a DFS on the tree to retrieve all paths
* @param tree
*/
public void runDFS(Integer node) {
if (getAdjacentsToNode(node) != null) {
for (Integer adjNode : getAdjacentsToNode(node)) {
currentPath.add(adjNode);
runDFS(adjNode);
}
currentPath.remove(currentPath.size() -1);
} else {
paths.add(currentPath.toArray(new Integer[0]));
currentPath.remove(currentPath.size() -1);
}
}
因此，整个班级看起来像这样：
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class DFS {
private List<Integer> currentPath = new ArrayList<Integer>();
private List<Integer[]> paths = new ArrayList<Integer[]>();
private ArrayList<Set<Integer>> tree;
private Integer root;
/**
* constructor
* @param tree
*/
public DFS(ArrayList<Set<Integer>> tree) {
this.tree = tree;
}
public List<Integer[]> getPaths() {
return paths;
}
public void setPaths(List<Integer[]> paths) {
this.paths = paths;
}
public Integer getRoot() {
return root;
}
public void setRoot(Integer root) {
this.root = root;
}
/**
* initializes our search, sets the root and adds it to the path
*/
public void initialize() {
root = -1;
for (Integer node : tree.get(0)) {
root = node;
}
currentPath.add(root);
}
/**
* This returns the adjacent nodes to an integer node in the tree
* @param node
* @return a set of adjacent nodes, and null otherwise
*/
public Set<Integer> getAdjacentsToNode(Integer node) {
for (int i = 0; i < tree.size(); i++) {
Set<Integer> level = tree.get(i);
if (level.contains(node) && i < tree.size() - 1) {
return tree.get(i + 1);
}
}
return null;
}
/**
* runs a DFS on the tree to retrieve all paths
* @param tree
*/
public void runDFS(Integer node) {
if (getAdjacentsToNode(node) != null) {
for (Integer adjNode : getAdjacentsToNode(node)) {
currentPath.add(adjNode);
runDFS(adjNode);
}
currentPath.remove(currentPath.size() -1);
} else {
paths.add(currentPath.toArray(new Integer[0]));
currentPath.remove(currentPath.size() -1);
}
}
}
为了测试它，我们试试这个：
public static void main(String[] args) {
ArrayList<Set<Integer>> tree = new ArrayList<Set<Integer>>();
Set<Integer> level1 = new HashSet<Integer>();
level1.add(new Integer(1));
Set<Integer> level2 = new HashSet<Integer>();
level2.add(new Integer(2));
level2.add(new Integer(3));
Set<Integer> level3 = new HashSet<Integer>();
level3.add(new Integer(4));
Set<Integer> level4 = new HashSet<Integer>();
level4.add(new Integer(5));
level4.add(new Integer(6));
level4.add(new Integer(7));
tree.add(level1);
tree.add(level2);
tree.add(level3);
tree.add(level4);
DFS dfsSearch = new DFS(tree);
dfsSearch.initialize();
dfsSearch.runDFS(dfsSearch.getRoot());
for (Integer[] path : dfsSearch.getPaths()) {
System.out.println(Arrays.toString(path));
}
我们得到以下输出：
[1, 2, 4, 5]
[1, 2, 4, 6]
[1, 2, 4, 7]
[1, 3, 4, 5]
[1, 3, 4, 6]
[1, 3, 4, 7]