java实现二叉树层次遍历
任意选定一棵至少含有8个结点的二叉树,编程实现:(1)按层次遍历算法遍历该二叉树,并给出遍历结果;(2)利用层次遍历算法显示所有叶结点。显示所有叶结点...
任意选定一棵至少含有8个结点的二叉树,编程实现:
(1)按层次遍历算法遍历该二叉树,并给出遍历结果;
(2)利用层次遍历算法显示所有叶结点。
显示所有叶结点 展开
(1)按层次遍历算法遍历该二叉树,并给出遍历结果;
(2)利用层次遍历算法显示所有叶结点。
显示所有叶结点 展开
1个回答
展开全部
import java.util.ArrayList;
public class TreeNode {
private TreeNode leftNode;
private TreeNode rightNode;
private String nodeName;
public TreeNode getLeftNode() {
return leftNode;
}
public void setLeftNode(TreeNode leftNode) {
this.leftNode = leftNode;
}
public TreeNode getRightNode() {
return rightNode;
}
public void setRightNode(TreeNode rightNode) {
this.rightNode = rightNode;
}
public String getNodeName() {
return nodeName;
}
public void setNodeName(String nodeName) {
this.nodeName = nodeName;
}
public static int level=0;
public static void findNodeByLevel(ArrayList<TreeNode> nodes){
if(nodes==null||nodes.size()==0){
return ;
}
level++;
ArrayList<TreeNode> temp = new ArrayList();
for(TreeNode node:nodes){
System.out.println("第"+level+"层:"+node.getNodeName());
if(node.getLeftNode()!=null){
temp.add(node.getLeftNode());
}
if(node.getRightNode()!=null){
temp.add(node.getRightNode());
}
}
nodes.removeAll(nodes);
findNodeByLevel(temp);
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root = new TreeNode();
root.setNodeName("root");
TreeNode node1 = new TreeNode();
node1.setNodeName("node1");
TreeNode node3 = new TreeNode();
node3.setNodeName("node3");
TreeNode node7 = new TreeNode();
node7.setNodeName("node7");
TreeNode node8 = new TreeNode();
node8.setNodeName("node8");
TreeNode node4 = new TreeNode();
node4.setNodeName("node4");
TreeNode node2 = new TreeNode();
node2.setNodeName("node2");
TreeNode node5 = new TreeNode();
node5.setNodeName("node5");
TreeNode node6 = new TreeNode();
node6.setNodeName("node6");
root.setLeftNode(node1);
node1.setLeftNode(node3);
node3.setLeftNode(node7);
node3.setRightNode(node8);
node1.setRightNode(node4);
root.setRightNode(node2);
node2.setLeftNode(node5);
node2.setRightNode(node6);
ArrayList<TreeNode> nodes = new ArrayList<TreeNode>();
nodes.add(root);
findNodeByLevel(nodes);
}
}
public class TreeNode {
private TreeNode leftNode;
private TreeNode rightNode;
private String nodeName;
public TreeNode getLeftNode() {
return leftNode;
}
public void setLeftNode(TreeNode leftNode) {
this.leftNode = leftNode;
}
public TreeNode getRightNode() {
return rightNode;
}
public void setRightNode(TreeNode rightNode) {
this.rightNode = rightNode;
}
public String getNodeName() {
return nodeName;
}
public void setNodeName(String nodeName) {
this.nodeName = nodeName;
}
public static int level=0;
public static void findNodeByLevel(ArrayList<TreeNode> nodes){
if(nodes==null||nodes.size()==0){
return ;
}
level++;
ArrayList<TreeNode> temp = new ArrayList();
for(TreeNode node:nodes){
System.out.println("第"+level+"层:"+node.getNodeName());
if(node.getLeftNode()!=null){
temp.add(node.getLeftNode());
}
if(node.getRightNode()!=null){
temp.add(node.getRightNode());
}
}
nodes.removeAll(nodes);
findNodeByLevel(temp);
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root = new TreeNode();
root.setNodeName("root");
TreeNode node1 = new TreeNode();
node1.setNodeName("node1");
TreeNode node3 = new TreeNode();
node3.setNodeName("node3");
TreeNode node7 = new TreeNode();
node7.setNodeName("node7");
TreeNode node8 = new TreeNode();
node8.setNodeName("node8");
TreeNode node4 = new TreeNode();
node4.setNodeName("node4");
TreeNode node2 = new TreeNode();
node2.setNodeName("node2");
TreeNode node5 = new TreeNode();
node5.setNodeName("node5");
TreeNode node6 = new TreeNode();
node6.setNodeName("node6");
root.setLeftNode(node1);
node1.setLeftNode(node3);
node3.setLeftNode(node7);
node3.setRightNode(node8);
node1.setRightNode(node4);
root.setRightNode(node2);
node2.setLeftNode(node5);
node2.setRightNode(node6);
ArrayList<TreeNode> nodes = new ArrayList<TreeNode>();
nodes.add(root);
findNodeByLevel(nodes);
}
}
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询