树的Breadth-First-Search和Depth-First-Search的java实现

假设待搜索的多叉树结构如下：

/ | \

3 5 7

/ | \ | / \

9 2 4 6 8 10

对应的代码如下：

package com.test.Tree;import java.util.*;public class N_aryTree {Tree tree=null;public static void main(String[] args) {// TODO Auto-generated method stubN_aryTree n_aryTree=new N_aryTree();}public N_aryTree() {// TODO Auto-generated constructor stubTree tree=new Tree();tree.BFS(tree.rootNode, 10);tree.DFS(tree.rootNode, 11);}}class Node{public int val;public List<Node> children;public int getVal() {return val;}public void setVal(int val) {this.val = val;}public List<Node> getChildren() {return this.children;}public void setChildren(List<Node> children) {this.children = children;}public Node(int val) {this.val = val;this.children = null;}}class Tree{Node rootNode;Node[] nodes=new Node[10]; List<Node> tempChildren=new Vector<Node>();List<Node> Children1=new Vector<Node>();List<Node> Children2=new Vector<Node>();List<Node> Children3=new Vector<Node>();List<Node> Children4=new Vector<Node>();Queue<Node> childrenQue=new LinkedList<Node>();Stack<Node> childrenSta=new Stack<Node>();public Tree() {// TODO Auto-generated constructor stub//先给每个节点赋值for (int i = 0; i < nodes.length; i++) {nodes[i]=new Node(i+1);}//建立节点之间的父子关系Children1.add(nodes[8]);        Children1.add(nodes[1]);Children1.add(nodes[3]);        nodes[2].setChildren(Children1);Children2.add(nodes[5]);nodes[4].setChildren(Children2);Children3.add(nodes[7]);Children3.add(nodes[9]);nodes[6].setChildren(Children3);Children4.add(nodes[2]);Children4.add(nodes[4]);Children4.add(nodes[6]);nodes[0].setChildren(Children4);rootNode=nodes[0];//System.out.println(nodes[2].children.size());}// breadth-first-search,node为待搜索的树的根节点，val为待搜索的节点的值public boolean BFS(Node node,int val) {System.out.println("BFS值为"+val+"的节点");if (node==null) {System.out.println("该树为空");return false;}      System.out.print("搜索路径：");if (node.getVal()==val) {return true;}childrenQue.offer(node);
/*      for (int i = 0; i < tempChildren.size(); i++) {childrenQue.offer(tempChildren.get(i));}*/while (!childrenQue.isEmpty()) {///System.out.println("hello");Node headNode=childrenQue.poll();System.out.print(headNode.getVal()+" ");if (headNode.getVal()==val) {System.out.println("找到节点");return true;}tempChildren=headNode.getChildren();if (tempChildren!=null) {for (int i = 0; i < tempChildren.size(); i++) {childrenQue.offer(tempChildren.get(i));}}
/*          for (int i = 0; i < tempChildren.size(); i++) {System.out.println("该节点孩子节点的个数为："+tempChildren.size()+" 第"+(i+1)+"个节点的值为："+tempChildren.get(i).val);childrenQue.offer(tempChildren.get(i));}*/}System.out.println("已遍历，该树不包含待寻节点");return false;}// depth-first-search,node为待搜索的树的根节点，val为待搜索的节点的值public boolean DFS(Node node,int val) {System.out.println("DFS值为"+val+"的节点");if (node==null) {System.out.println("该树为空");return false;}System.out.print("搜索路径：");if (node.getVal()==val) {return true;}childrenSta.push(node);System.out.print(node.val+" ");while (!childrenSta.isEmpty()) {Node headNode=childrenSta.peek();if (headNode.getChildren()==null||headNode.getChildren().isEmpty()) {Node toRmNode=childrenSta.pop();if (!childrenSta.isEmpty()) {Node fatherNode=childrenSta.peek();fatherNode.getChildren().remove(toRmNode);}}else {Node toAddNode=headNode.getChildren().get(0);System.out.print(toAddNode.getVal()+" ");if (toAddNode.getVal()==val) {System.out.println("找到节点");return true;}childrenSta.push(toAddNode);}}System.out.println("已遍历，该树不包含待寻节点");return false;}}

运行结果：

BFS值为10的节点
搜索路径：1 3 5 7 9 2 4 6 8 10 找到节点
DFS值为11的节点
搜索路径：1 3 9 2 4 5 6 7 8 10 已遍历，该树不包含待寻节点

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