我如何在Java中打印一个二叉树,这样输出就像:
4
/ \
2 5
我的节点:
public class Node<A extends Comparable> {
Node<A> left, right;
A data;
public Node(A data){
this.data = data;
}
}
我如何在Java中打印一个二叉树,这样输出就像:
4
/ \
2 5
我的节点:
public class Node<A extends Comparable> {
Node<A> left, right;
A data;
public Node(A data){
this.data = data;
}
}
当前回答
这是一个非常多功能的树打印机。不是最好看的,但能处理很多案子。如果你能弄清楚,可以随意添加斜杠。
package com.tomac120.NodePrinter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
/**
* Created by elijah on 6/28/16.
*/
public class NodePrinter{
final private List<List<PrintableNodePosition>> nodesByRow;
int maxColumnsLeft = 0;
int maxColumnsRight = 0;
int maxTitleLength = 0;
String sep = " ";
int depth = 0;
public NodePrinter(PrintableNode rootNode, int chars_per_node){
this.setDepth(rootNode,1);
nodesByRow = new ArrayList<>(depth);
this.addNode(rootNode._getPrintableNodeInfo(),0,0);
for (int i = 0;i<chars_per_node;i++){
//sep += " ";
}
}
private void setDepth(PrintableNode info, int depth){
if (depth > this.depth){
this.depth = depth;
}
if (info._getLeftChild() != null){
this.setDepth(info._getLeftChild(),depth+1);
}
if (info._getRightChild() != null){
this.setDepth(info._getRightChild(),depth+1);
}
}
private void addNode(PrintableNodeInfo node, int level, int position){
if (position < 0 && -position > maxColumnsLeft){
maxColumnsLeft = -position;
}
if (position > 0 && position > maxColumnsRight){
maxColumnsRight = position;
}
if (node.getTitleLength() > maxTitleLength){
maxTitleLength = node.getTitleLength();
}
List<PrintableNodePosition> row = this.getRow(level);
row.add(new PrintableNodePosition(node, level, position));
level++;
int depthToUse = Math.min(depth,6);
int levelToUse = Math.min(level,6);
int offset = depthToUse - levelToUse-1;
offset = (int)(Math.pow(offset,Math.log(depthToUse)*1.4));
offset = Math.max(offset,3);
PrintableNodeInfo leftChild = node.getLeftChildInfo();
PrintableNodeInfo rightChild = node.getRightChildInfo();
if (leftChild != null){
this.addNode(leftChild,level,position-offset);
}
if (rightChild != null){
this.addNode(rightChild,level,position+offset);
}
}
private List<PrintableNodePosition> getRow(int row){
if (row > nodesByRow.size() - 1){
nodesByRow.add(new LinkedList<>());
}
return nodesByRow.get(row);
}
public void print(){
int max_chars = this.maxColumnsLeft+maxColumnsRight+1;
int level = 0;
String node_format = "%-"+this.maxTitleLength+"s";
for (List<PrintableNodePosition> pos_arr : this.nodesByRow){
String[] chars = this.getCharactersArray(pos_arr,max_chars);
String line = "";
int empty_chars = 0;
for (int i=0;i<chars.length+1;i++){
String value_i = i < chars.length ? chars[i]:null;
if (chars.length + 1 == i || value_i != null){
if (empty_chars > 0) {
System.out.print(String.format("%-" + empty_chars + "s", " "));
}
if (value_i != null){
System.out.print(String.format(node_format,value_i));
empty_chars = -1;
} else{
empty_chars = 0;
}
} else {
empty_chars++;
}
}
System.out.print("\n");
int depthToUse = Math.min(6,depth);
int line_offset = depthToUse - level;
line_offset *= 0.5;
line_offset = Math.max(0,line_offset);
for (int i=0;i<line_offset;i++){
System.out.println("");
}
level++;
}
}
private String[] getCharactersArray(List<PrintableNodePosition> nodes, int max_chars){
String[] positions = new String[max_chars+1];
for (PrintableNodePosition a : nodes){
int pos_i = maxColumnsLeft + a.column;
String title_i = a.nodeInfo.getTitleFormatted(this.maxTitleLength);
positions[pos_i] = title_i;
}
return positions;
}
}
NodeInfo类
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public class PrintableNodeInfo {
public enum CLI_PRINT_COLOR {
RESET("\u001B[0m"),
BLACK("\u001B[30m"),
RED("\u001B[31m"),
GREEN("\u001B[32m"),
YELLOW("\u001B[33m"),
BLUE("\u001B[34m"),
PURPLE("\u001B[35m"),
CYAN("\u001B[36m"),
WHITE("\u001B[37m");
final String value;
CLI_PRINT_COLOR(String value){
this.value = value;
}
@Override
public String toString() {
return value;
}
}
private final String title;
private final PrintableNode leftChild;
private final PrintableNode rightChild;
private final CLI_PRINT_COLOR textColor;
public PrintableNodeInfo(String title, PrintableNode leftChild, PrintableNode rightChild){
this(title,leftChild,rightChild,CLI_PRINT_COLOR.BLACK);
}
public PrintableNodeInfo(String title, PrintableNode leftChild, PrintableNode righthild, CLI_PRINT_COLOR textColor){
this.title = title;
this.leftChild = leftChild;
this.rightChild = righthild;
this.textColor = textColor;
}
public String getTitle(){
return title;
}
public CLI_PRINT_COLOR getTextColor(){
return textColor;
}
public String getTitleFormatted(int max_chars){
return this.textColor+title+CLI_PRINT_COLOR.RESET;
/*
String title = this.title.length() > max_chars ? this.title.substring(0,max_chars+1):this.title;
boolean left = true;
while(title.length() < max_chars){
if (left){
title = " "+title;
} else {
title = title + " ";
}
}
return this.textColor+title+CLI_PRINT_COLOR.RESET;*/
}
public int getTitleLength(){
return title.length();
}
public PrintableNodeInfo getLeftChildInfo(){
if (leftChild == null){
return null;
}
return leftChild._getPrintableNodeInfo();
}
public PrintableNodeInfo getRightChildInfo(){
if (rightChild == null){
return null;
}
return rightChild._getPrintableNodeInfo();
}
}
NodePosition类
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public class PrintableNodePosition implements Comparable<PrintableNodePosition> {
public final int row;
public final int column;
public final PrintableNodeInfo nodeInfo;
public PrintableNodePosition(PrintableNodeInfo nodeInfo, int row, int column){
this.row = row;
this.column = column;
this.nodeInfo = nodeInfo;
}
@Override
public int compareTo(PrintableNodePosition o) {
return Integer.compare(this.column,o.column);
}
}
最后是节点接口
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public interface PrintableNode {
PrintableNodeInfo _getPrintableNodeInfo();
PrintableNode _getLeftChild();
PrintableNode _getRightChild();
}
其他回答
一个Scala解决方案,改编自Vasya Novikov的答案,专门用于二叉树:
/** An immutable Binary Tree. */
case class BTree[T](value: T, left: Option[BTree[T]], right: Option[BTree[T]]) {
/* Adapted from: http://stackoverflow.com/a/8948691/643684 */
def pretty: String = {
def work(tree: BTree[T], prefix: String, isTail: Boolean): String = {
val (line, bar) = if (isTail) ("└── ", " ") else ("├── ", "│")
val curr = s"${prefix}${line}${tree.value}"
val rights = tree.right match {
case None => s"${prefix}${bar} ├── ∅"
case Some(r) => work(r, s"${prefix}${bar} ", false)
}
val lefts = tree.left match {
case None => s"${prefix}${bar} └── ∅"
case Some(l) => work(l, s"${prefix}${bar} ", true)
}
s"${curr}\n${rights}\n${lefts}"
}
work(this, "", true)
}
}
这是打印树的一个非常简单的解决方案。它不是那么漂亮,但它真的很简单:
enum { kWidth = 6 };
void PrintSpace(int n)
{
for (int i = 0; i < n; ++i)
printf(" ");
}
void PrintTree(struct Node * root, int level)
{
if (!root) return;
PrintTree(root->right, level + 1);
PrintSpace(level * kWidth);
printf("%d", root->data);
PrintTree(root->left, level + 1);
}
样例输出:
106 105 104 103 102 101 100
试试这个:
public static void print(int[] minHeap, int minWidth) {
int size = minHeap.length;
int level = log2(size);
int maxLength = (int) Math.pow(2, level) * minWidth;
int currentLevel = -1 ;
int width = maxLength;
for (int i = 0; i < size; i++) {
if (log2(i + 1) > currentLevel) {
currentLevel++;
System.out.println();
width = maxLength / (int) Math.pow(2, currentLevel);
}
System.out.print(StringUtils.center(String.valueOf(minHeap[i]), width));
}
System.out.println();
}
private static int log2(int n) {
return (int) (Math.log(n) / Math.log(2));
}
这段代码片段的思想是用maxLength(即底线的长度)除以每一行的元素数量来得到块宽度。然后把元素放在每个块的中间。
参数minWidth表示底部行中块的长度。
用一张图片来说明想法并展示结果。
下面是可视化树的另一种方法:将节点保存为xml文件,然后让浏览器显示层次结构:
class treeNode{
int key;
treeNode left;
treeNode right;
public treeNode(int key){
this.key = key;
left = right = null;
}
public void printNode(StringBuilder output, String dir){
output.append("<node key='" + key + "' dir='" + dir + "'>");
if(left != null)
left.printNode(output, "l");
if(right != null)
right.printNode(output, "r");
output.append("</node>");
}
}
class tree{
private treeNode treeRoot;
public tree(int key){
treeRoot = new treeNode(key);
}
public void insert(int key){
insert(treeRoot, key);
}
private treeNode insert(treeNode root, int key){
if(root == null){
treeNode child = new treeNode(key);
return child;
}
if(key < root.key)
root.left = insert(root.left, key);
else if(key > root.key)
root.right = insert(root.right, key);
return root;
}
public void saveTreeAsXml(){
StringBuilder strOutput = new StringBuilder();
strOutput.append("<?xml version=\"1.0\" encoding=\"UTF-8\"?>");
treeRoot.printNode(strOutput, "root");
try {
PrintWriter writer = new PrintWriter("C:/tree.xml", "UTF-8");
writer.write(strOutput.toString());
writer.close();
}
catch (FileNotFoundException e){
}
catch(UnsupportedEncodingException e){
}
}
}
下面是测试它的代码:
tree t = new tree(1);
t.insert(10);
t.insert(5);
t.insert(4);
t.insert(20);
t.insert(40);
t.insert(30);
t.insert(80);
t.insert(60);
t.insert(50);
t.saveTreeAsXml();
输出如下所示:
迈克尔。克鲁兹曼,我不得不说,这人不错。这很有用。
然而,上面的方法只适用于个位数:如果您要使用多个数字,结构将会错位,因为您使用的是空格而不是制表符。
至于我后来的代码,我需要更多的数字,所以我自己编写了一个程序。
它现在有一些bug,现在我感觉很懒去纠正它们,但它打印得非常漂亮,节点可以接受更大数量的数字。
这棵树不会像问题提到的那样,但它旋转了270度:)
public static void printBinaryTree(TreeNode root, int level){
if(root==null)
return;
printBinaryTree(root.right, level+1);
if(level!=0){
for(int i=0;i<level-1;i++)
System.out.print("|\t");
System.out.println("|-------"+root.val);
}
else
System.out.println(root.val);
printBinaryTree(root.left, level+1);
}
将此函数与您自己指定的TreeNode一起放置,并保持初始级别为0,并享受!
以下是一些输出示例:
| | |-------11
| |-------10
| | |-------9
|-------8
| | |-------7
| |-------6
| | |-------5
4
| |-------3
|-------2
| |-------1
| | | |-------10
| | |-------9
| |-------8
| | |-------7
|-------6
| |-------5
4
| |-------3
|-------2
| |-------1
唯一的问题是延伸的分支;我会尽快解决这个问题,但在此之前你也可以使用它。