如何在Java中打印二叉树,使输出类似于:
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;
}
}
如何在Java中打印二叉树,使输出类似于:
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;
}
}
Answers:
我创建了简单的二叉树打印机。您可以根据需要使用和修改它,但是仍然没有对其进行优化。我认为这里有很多事情可以改进;)
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class BTreePrinterTest {
private static Node<Integer> test1() {
Node<Integer> root = new Node<Integer>(2);
Node<Integer> n11 = new Node<Integer>(7);
Node<Integer> n12 = new Node<Integer>(5);
Node<Integer> n21 = new Node<Integer>(2);
Node<Integer> n22 = new Node<Integer>(6);
Node<Integer> n23 = new Node<Integer>(3);
Node<Integer> n24 = new Node<Integer>(6);
Node<Integer> n31 = new Node<Integer>(5);
Node<Integer> n32 = new Node<Integer>(8);
Node<Integer> n33 = new Node<Integer>(4);
Node<Integer> n34 = new Node<Integer>(5);
Node<Integer> n35 = new Node<Integer>(8);
Node<Integer> n36 = new Node<Integer>(4);
Node<Integer> n37 = new Node<Integer>(5);
Node<Integer> n38 = new Node<Integer>(8);
root.left = n11;
root.right = n12;
n11.left = n21;
n11.right = n22;
n12.left = n23;
n12.right = n24;
n21.left = n31;
n21.right = n32;
n22.left = n33;
n22.right = n34;
n23.left = n35;
n23.right = n36;
n24.left = n37;
n24.right = n38;
return root;
}
private static Node<Integer> test2() {
Node<Integer> root = new Node<Integer>(2);
Node<Integer> n11 = new Node<Integer>(7);
Node<Integer> n12 = new Node<Integer>(5);
Node<Integer> n21 = new Node<Integer>(2);
Node<Integer> n22 = new Node<Integer>(6);
Node<Integer> n23 = new Node<Integer>(9);
Node<Integer> n31 = new Node<Integer>(5);
Node<Integer> n32 = new Node<Integer>(8);
Node<Integer> n33 = new Node<Integer>(4);
root.left = n11;
root.right = n12;
n11.left = n21;
n11.right = n22;
n12.right = n23;
n22.left = n31;
n22.right = n32;
n23.left = n33;
return root;
}
public static void main(String[] args) {
BTreePrinter.printNode(test1());
BTreePrinter.printNode(test2());
}
}
class Node<T extends Comparable<?>> {
Node<T> left, right;
T data;
public Node(T data) {
this.data = data;
}
}
class BTreePrinter {
public static <T extends Comparable<?>> void printNode(Node<T> root) {
int maxLevel = BTreePrinter.maxLevel(root);
printNodeInternal(Collections.singletonList(root), 1, maxLevel);
}
private static <T extends Comparable<?>> void printNodeInternal(List<Node<T>> nodes, int level, int maxLevel) {
if (nodes.isEmpty() || BTreePrinter.isAllElementsNull(nodes))
return;
int floor = maxLevel - level;
int endgeLines = (int) Math.pow(2, (Math.max(floor - 1, 0)));
int firstSpaces = (int) Math.pow(2, (floor)) - 1;
int betweenSpaces = (int) Math.pow(2, (floor + 1)) - 1;
BTreePrinter.printWhitespaces(firstSpaces);
List<Node<T>> newNodes = new ArrayList<Node<T>>();
for (Node<T> node : nodes) {
if (node != null) {
System.out.print(node.data);
newNodes.add(node.left);
newNodes.add(node.right);
} else {
newNodes.add(null);
newNodes.add(null);
System.out.print(" ");
}
BTreePrinter.printWhitespaces(betweenSpaces);
}
System.out.println("");
for (int i = 1; i <= endgeLines; i++) {
for (int j = 0; j < nodes.size(); j++) {
BTreePrinter.printWhitespaces(firstSpaces - i);
if (nodes.get(j) == null) {
BTreePrinter.printWhitespaces(endgeLines + endgeLines + i + 1);
continue;
}
if (nodes.get(j).left != null)
System.out.print("/");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(i + i - 1);
if (nodes.get(j).right != null)
System.out.print("\\");
else
BTreePrinter.printWhitespaces(1);
BTreePrinter.printWhitespaces(endgeLines + endgeLines - i);
}
System.out.println("");
}
printNodeInternal(newNodes, level + 1, maxLevel);
}
private static void printWhitespaces(int count) {
for (int i = 0; i < count; i++)
System.out.print(" ");
}
private static <T extends Comparable<?>> int maxLevel(Node<T> node) {
if (node == null)
return 0;
return Math.max(BTreePrinter.maxLevel(node.left), BTreePrinter.maxLevel(node.right)) + 1;
}
private static <T> boolean isAllElementsNull(List<T> list) {
for (Object object : list) {
if (object != null)
return false;
}
return true;
}
}
输出1:
2
/ \
/ \
/ \
/ \
7 5
/ \ / \
/ \ / \
2 6 3 6
/ \ / \ / \ / \
5 8 4 5 8 4 5 8
输出2:
2
/ \
/ \
/ \
/ \
7 5
/ \ \
/ \ \
2 6 9
/ \ /
5 8 4
按行打印[大]树。
输出示例:
z
├── c
│ ├── a
│ └── b
├── d
├── e
│ └── asdf
└── f
码:
public class TreeNode {
final String name;
final List<TreeNode> children;
public TreeNode(String name, List<TreeNode> children) {
this.name = name;
this.children = children;
}
public String toString() {
StringBuilder buffer = new StringBuilder(50);
print(buffer, "", "");
return buffer.toString();
}
private void print(StringBuilder buffer, String prefix, String childrenPrefix) {
buffer.append(prefix);
buffer.append(name);
buffer.append('\n');
for (Iterator<TreeNode> it = children.iterator(); it.hasNext();) {
TreeNode next = it.next();
if (it.hasNext()) {
next.print(buffer, childrenPrefix + "├── ", childrenPrefix + "│ ");
} else {
next.print(buffer, childrenPrefix + "└── ", childrenPrefix + " ");
}
}
}
}
PS:这个答案并不完全集中在“二叉树”上,而是打印各种树。解决方案的灵感来自Linux中的“ tree”命令。
children.get(children.size() - 1)
如果将HashMap用于儿童,您将如何重写?除了这一部分,我设法修改了其他所有部分。
HashMap<String, List<String>>
?
HashMap<String, Node>
。字符串是节点的ID。
我为此做了一个改进的算法,可以很好地处理不同大小的节点。它使用线条自上而下打印。
package alg;
import java.util.ArrayList;
import java.util.List;
/**
* Binary tree printer
*
* @author MightyPork
*/
public class TreePrinter
{
/** Node that can be printed */
public interface PrintableNode
{
/** Get left child */
PrintableNode getLeft();
/** Get right child */
PrintableNode getRight();
/** Get text to be printed */
String getText();
}
/**
* Print a tree
*
* @param root
* tree root node
*/
public static void print(PrintableNode root)
{
List<List<String>> lines = new ArrayList<List<String>>();
List<PrintableNode> level = new ArrayList<PrintableNode>();
List<PrintableNode> next = new ArrayList<PrintableNode>();
level.add(root);
int nn = 1;
int widest = 0;
while (nn != 0) {
List<String> line = new ArrayList<String>();
nn = 0;
for (PrintableNode n : level) {
if (n == null) {
line.add(null);
next.add(null);
next.add(null);
} else {
String aa = n.getText();
line.add(aa);
if (aa.length() > widest) widest = aa.length();
next.add(n.getLeft());
next.add(n.getRight());
if (n.getLeft() != null) nn++;
if (n.getRight() != null) nn++;
}
}
if (widest % 2 == 1) widest++;
lines.add(line);
List<PrintableNode> tmp = level;
level = next;
next = tmp;
next.clear();
}
int perpiece = lines.get(lines.size() - 1).size() * (widest + 4);
for (int i = 0; i < lines.size(); i++) {
List<String> line = lines.get(i);
int hpw = (int) Math.floor(perpiece / 2f) - 1;
if (i > 0) {
for (int j = 0; j < line.size(); j++) {
// split node
char c = ' ';
if (j % 2 == 1) {
if (line.get(j - 1) != null) {
c = (line.get(j) != null) ? '┴' : '┘';
} else {
if (j < line.size() && line.get(j) != null) c = '└';
}
}
System.out.print(c);
// lines and spaces
if (line.get(j) == null) {
for (int k = 0; k < perpiece - 1; k++) {
System.out.print(" ");
}
} else {
for (int k = 0; k < hpw; k++) {
System.out.print(j % 2 == 0 ? " " : "─");
}
System.out.print(j % 2 == 0 ? "┌" : "┐");
for (int k = 0; k < hpw; k++) {
System.out.print(j % 2 == 0 ? "─" : " ");
}
}
}
System.out.println();
}
// print line of numbers
for (int j = 0; j < line.size(); j++) {
String f = line.get(j);
if (f == null) f = "";
int gap1 = (int) Math.ceil(perpiece / 2f - f.length() / 2f);
int gap2 = (int) Math.floor(perpiece / 2f - f.length() / 2f);
// a number
for (int k = 0; k < gap1; k++) {
System.out.print(" ");
}
System.out.print(f);
for (int k = 0; k < gap2; k++) {
System.out.print(" ");
}
}
System.out.println();
perpiece /= 2;
}
}
}
要将其用于Tree,让您的Node
类实现PrintableNode
。
输出示例:
2952:0
┌───────────────────────┴───────────────────────┐
1249:-1 5866:0
┌───────────┴───────────┐ ┌───────────┴───────────┐
491:-1 1572:0 4786:1 6190:0
┌─────┘ └─────┐ ┌─────┴─────┐
339:0 5717:0 6061:0 6271:0
???????????
不是节点之间的线,而应该只是一些UTF8 ans问题。无论如何,很棒的东西,我不得不说。对我来说最好的答案,因为它确实很容易使用。
public static class Node<T extends Comparable<T>> {
T value;
Node<T> left, right;
public void insertToTree(T v) {
if (value == null) {
value = v;
return;
}
if (v.compareTo(value) < 0) {
if (left == null) {
left = new Node<T>();
}
left.insertToTree(v);
} else {
if (right == null) {
right = new Node<T>();
}
right.insertToTree(v);
}
}
public void printTree(OutputStreamWriter out) throws IOException {
if (right != null) {
right.printTree(out, true, "");
}
printNodeValue(out);
if (left != null) {
left.printTree(out, false, "");
}
}
private void printNodeValue(OutputStreamWriter out) throws IOException {
if (value == null) {
out.write("<null>");
} else {
out.write(value.toString());
}
out.write('\n');
}
// use string and not stringbuffer on purpose as we need to change the indent at each recursion
private void printTree(OutputStreamWriter out, boolean isRight, String indent) throws IOException {
if (right != null) {
right.printTree(out, true, indent + (isRight ? " " : " | "));
}
out.write(indent);
if (isRight) {
out.write(" /");
} else {
out.write(" \\");
}
out.write("----- ");
printNodeValue(out);
if (left != null) {
left.printTree(out, false, indent + (isRight ? " | " : " "));
}
}
}
将打印:
/----- 20
| \----- 15
/----- 14
| \----- 13
/----- 12
| | /----- 11
| \----- 10
| \----- 9
8
| /----- 7
| /----- 6
| | \----- 5
\----- 4
| /----- 3
\----- 2
\----- 1
用于输入
8 4 12 2 6 10 14 1 3 5 7 9 11 13 20 15
这是@anurag的答案的变体-烦我看看多余的| s
改编自Vasya Novikov的答案,使其更加二进制,并使用a StringBuilder
表示效率(String
在Java中将对象连接在一起通常效率不高)。
public StringBuilder toString(StringBuilder prefix, boolean isTail, StringBuilder sb) {
if(right!=null) {
right.toString(new StringBuilder().append(prefix).append(isTail ? "│ " : " "), false, sb);
}
sb.append(prefix).append(isTail ? "└── " : "┌── ").append(value.toString()).append("\n");
if(left!=null) {
left.toString(new StringBuilder().append(prefix).append(isTail ? " " : "│ "), true, sb);
}
return sb;
}
@Override
public String toString() {
return this.toString(new StringBuilder(), true, new StringBuilder()).toString();
}
输出:
│ ┌── 7
│ ┌── 6
│ │ └── 5
└── 4
│ ┌── 3
└── 2
└── 1
└── 0
right != null
我进行编辑并对其进行测试时,isTail的值应为false ,它可以正常工作。
michal.kreuzman不错,我不得不说。
当我发现这确实对我有帮助时,我懒得自己编写程序并在网上搜索代码。
但是我恐怕会看到它仅适用于一位数字,就好像您将使用一个以上的数字一样,因为您使用的是空格而不是制表符,所以结构会放错位置,并且程序将失去其使用。
至于我以后的代码,我需要一些更大的输入(至少超过10个),这对我来说不起作用,在找不到任何东西的情况下在网上搜索了很多之后,我自己编写了一个程序。
它现在有一些错误,现在我又懒得去纠正它们,但是它打印的非常漂亮,节点可以取任何大的价值。
该树不会像问题中提到的那样旋转,但会旋转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放置在一起,并将级别初始保持为0,然后尽情享受吧!
以下是一些示例输出:
| | |-------11
| |-------10
| | |-------9
|-------8
| | |-------7
| |-------6
| | |-------5
4
| |-------3
|-------2
| |-------1
| | | |-------10
| | |-------9
| |-------8
| | |-------7
|-------6
| |-------5
4
| |-------3
|-------2
| |-------1
唯一的问题是分支的扩展。我将尝试尽快解决该问题,但在此之前您也可以使用它。
您的树将需要为每一层增加两倍的距离:
一个 / \ / \ / \ / \ 公元前 /// /// 定义 / / / \ / \ / \ 希克尔姆诺
您可以将树保存为一个数组数组,每个深度一个数组:
[[a],[b,c],[d,e,f,g],[h,i,j,k,l,m,n,o]]
如果树未满,则需要在该数组中包括空值:
一个 / \ / \ / \ / \ 公元前 /// /// 定义 / / \ / \ \ 希克尔莫 [[a],[b,c],[d,e,f,g],[h,i,,k,l,m,,o]]
然后,您可以遍历数组以打印树,根据深度在第一个元素之前和元素之间打印空格,并根据是否填充下一层的数组中的相应元素来打印线条。如果值的长度可以超过一个字符,则需要在创建数组表示形式时找到最长的值,并相应地乘以所有宽度和行数。
我发现VasyaNovikov的答案对于打印大型通用树非常有用,并将其修改为二叉树
码:
class TreeNode {
Integer data = null;
TreeNode left = null;
TreeNode right = null;
TreeNode(Integer data) {this.data = data;}
public void print() {
print("", this, false);
}
public void print(String prefix, TreeNode n, boolean isLeft) {
if (n != null) {
System.out.println (prefix + (isLeft ? "|-- " : "\\-- ") + n.data);
print(prefix + (isLeft ? "| " : " "), n.left, true);
print(prefix + (isLeft ? "| " : " "), n.right, false);
}
}
}
样本输出:
\-- 7
|-- 3
| |-- 1
| | \-- 2
| \-- 5
| |-- 4
| \-- 6
\-- 11
|-- 9
| |-- 8
| \-- 10
\-- 13
|-- 12
\-- 14
Scala语言的解决方案,类似于我在java中编写的内容:
case class Node(name: String, children: Node*) {
def toTree: String = toTree("", "").mkString("\n")
private def toTree(prefix: String, childrenPrefix: String): Seq[String] = {
val firstLine = prefix + this.name
val firstChildren = this.children.dropRight(1).flatMap { child =>
child.toTree(childrenPrefix + "├── ", childrenPrefix + "│ ")
}
val lastChild = this.children.takeRight(1).flatMap { child =>
child.toTree(childrenPrefix + "└── ", childrenPrefix + " ")
}
firstLine +: firstChildren ++: lastChild
}
}
输出示例:
vasya
├── frosya
│ ├── petya
│ │ └── masha
│ └── kolya
└── frosya2
我知道你们都有很好的解决方案。我只想分享我的照片-也许那不是最好的方法,但是对我自己来说是完美的!
随着python
与pip
上,这是很简单的!繁荣!
在Mac或Ubuntu上(我的是Mac)
$ pip install drawtree
$python
,进入python控制台;你可以用其他方式做到from drawtree import draw_level_order
draw_level_order('{2,1,3,0,7,9,1,2,#,1,0,#,#,8,8,#,#,#,#,7}')
完成!
2
/ \
/ \
/ \
1 3
/ \ / \
0 7 9 1
/ / \ / \
2 1 0 8 8
/
7
来源追踪:
在我看到这篇文章之前,我去过Google“二叉树纯文本”
我发现了这个https://www.reddit.com/r/learnpython/comments/3naiq8/draw_binary_tree_in_plain_text/,将我定向到此https://github.com/msbanik/drawtree
java
他期望的Java答案标记问题时,:)
public void printPreety() {
List<TreeNode> list = new ArrayList<TreeNode>();
list.add(head);
printTree(list, getHeight(head));
}
public int getHeight(TreeNode head) {
if (head == null) {
return 0;
} else {
return 1 + Math.max(getHeight(head.left), getHeight(head.right));
}
}
/**
* pass head node in list and height of the tree
*
* @param levelNodes
* @param level
*/
private void printTree(List<TreeNode> levelNodes, int level) {
List<TreeNode> nodes = new ArrayList<TreeNode>();
//indentation for first node in given level
printIndentForLevel(level);
for (TreeNode treeNode : levelNodes) {
//print node data
System.out.print(treeNode == null?" ":treeNode.data);
//spacing between nodes
printSpacingBetweenNodes(level);
//if its not a leaf node
if(level>1){
nodes.add(treeNode == null? null:treeNode.left);
nodes.add(treeNode == null? null:treeNode.right);
}
}
System.out.println();
if(level>1){
printTree(nodes, level-1);
}
}
private void printIndentForLevel(int level){
for (int i = (int) (Math.pow(2,level-1)); i >0; i--) {
System.out.print(" ");
}
}
private void printSpacingBetweenNodes(int level){
//spacing between nodes
for (int i = (int) ((Math.pow(2,level-1))*2)-1; i >0; i--) {
System.out.print(" ");
}
}
Prints Tree in following format:
4
3 7
1 5 8
2 10
9
这是打印出树的非常简单的解决方案。它不是那么漂亮,但确实很简单:
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);
}
样本输出:
106 105 104 103 102 101 100
基于VasyaNovikov的答案。通过Java魔术进行了改进:泛型和功能接口。
/**
* Print a tree structure in a pretty ASCII fromat.
* @param prefix Currnet previx. Use "" in initial call!
* @param node The current node. Pass the root node of your tree in initial call.
* @param getChildrenFunc A {@link Function} that returns the children of a given node.
* @param isTail Is node the last of its sibblings. Use true in initial call. (This is needed for pretty printing.)
* @param <T> The type of your nodes. Anything that has a toString can be used.
*/
private <T> void printTreeRec(String prefix, T node, Function<T, List<T>> getChildrenFunc, boolean isTail) {
String nodeName = node.toString();
String nodeConnection = isTail ? "└── " : "├── ";
log.debug(prefix + nodeConnection + nodeName);
List<T> children = getChildrenFunc.apply(node);
for (int i = 0; i < children.size(); i++) {
String newPrefix = prefix + (isTail ? " " : "│ ");
printTreeRec(newPrefix, children.get(i), getChildrenFunc, i == children.size()-1);
}
}
初始调用示例:
Function<ChecksumModel, List<ChecksumModel>> getChildrenFunc = node -> getChildrenOf(node)
printTreeRec("", rootNode, getChildrenFunc, true);
将输出类似
└── rootNode
├── childNode1
├── childNode2
│ ├── childNode2.1
│ ├── childNode2.2
│ └── childNode2.3
├── childNode3
└── childNode4
我用Java编写了一个二叉树打印机。
代码在GitHub上。
它尚未针对运行时效率进行优化,但是由于我们正在谈论使用ASCII进行打印,因此我认为它不会在非常大的树上使用。它确实具有一些不错的功能。
包括一些演示/测试程序。
由程序打印的随机生成的二叉树的示例如下。这说明了空间的有效利用,其中一个大的右子树在一个小的左子树下延伸:
seven
/ \
/ \
/ \
/ \
/ \
/ \
five thirteen
/ \ / \
/ \ / \
/ \ / \
three six / \
/ \ / \
/ \ / \
one four / \
\ / \
two / \
nine twenty four
/ \ / \
/ \ / \
/ \ / \
eight twelve / \
/ / \
ten / \
\ / \
eleven / \
/ \
/ \
/ \
eighteen twenty seven
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ twenty five twenty eight
/ \ \ \
/ \ twenty six thirty
fourteen nineteen /
\ \ twenty nine
sixteen twenty three
/ \ /
/ \ twenty two
/ \ /
/ \ twenty
/ \ \
fifteen seventeen twenty one
在页面上打印所有五个节点二叉树(带有顺序标签)的示例:
one one one one one one one
\ \ \ \ \ \ \
two two two two two three three
\ \ \ \ \ / \ / \
three three four five five two four two five
\ \ / \ / / \ /
four five / \ three four five four
\ / three five \ /
five four four three
one one one one one one one two
\ \ \ \ \ \ \ / \
four four five five five five five / \
/ \ / \ / / / / / one three
two five / \ two two three four four \
\ three five \ \ / \ / / four
three / three four two four two three \
two \ / \ / five
four three three two
two two two two three three three
/ \ / \ / \ / \ / \ / \ / \
/ \ one four one five one five one four / \ two four
one three / \ / / \ \ / \ / \
\ / \ three four two five one five one five
five three five \ / \ /
/ four three two four
four
three four four four four four five
/ \ / \ / \ / \ / \ / \ /
two five one five one five two five / \ / \ one
/ / \ \ / \ three five three five \
one four two three / \ / / two
\ / one three one two \
three two \ / three
two one \
four
five five five five five five five five
/ / / / / / / /
one one one one two two three three
\ \ \ \ / \ / \ / \ / \
two three four four / \ one four one four two four
\ / \ / / one three / \ /
four two four two three \ three two one
/ \ / four
three three two
five five five five five
/ / / / /
four four four four four
/ / / / /
one one two three three
\ \ / \ / /
two three / \ one two
\ / one three \ /
three two two one
以下是同一棵树以4种不同方式打印的示例,水平间距为1和3,对角和水平分支。
27
┌─────┴─────┐
13 29
┌──────┴──────┐ ┌─┴─┐
8 23 28 30
┌──┴──┐ ┌──┴──┐
4 11 21 26
┌─┴─┐ ┌┴┐ ┌─┴─┐ ┌┘
2 5 9 12 18 22 24
┌┴┐ └┐ └┐ ┌─┴─┐ └┐
1 3 6 10 17 19 25
└┐ ┌┘ └┐
7 15 20
┌─┴─┐
14 16
27
/ \
/ \
13 29
/ \ / \
/ \ 28 30
/ \
/ \
/ \
/ \
8 23
/ \ / \
/ \ / \
4 11 / \
/ \ / \ 21 26
2 5 9 12 / \ /
/ \ \ \ 18 22 24
1 3 6 10 / \ \
\ 17 19 25
7 / \
15 20
/ \
14 16
27
┌────────┴────────┐
13 29
┌─────────┴─────────┐ ┌──┴──┐
8 23 28 30
┌────┴────┐ ┌────┴────┐
4 11 21 26
┌──┴──┐ ┌─┴─┐ ┌──┴──┐ ┌┘
2 5 9 12 18 22 24
┌─┴─┐ └┐ └┐ ┌──┴──┐ └┐
1 3 6 10 17 19 25
└┐ ┌┘ └┐
7 15 20
┌──┴──┐
14 16
27
/ \
/ \
/ \
/ \
13 29
/ \ / \
/ \ / \
/ \ 28 30
/ \
/ \
/ \
/ \
/ \
8 23
/ \ / \
/ \ / \
/ \ / \
4 11 / \
/ \ / \ 21 26
2 5 9 12 / \ /
/ \ \ \ / \ 24
1 3 6 10 18 22 \
\ / \ 25
7 / \
17 19
/ \
15 20
/ \
/ \
14 16
@Makyen
注释来对我进行ping操作。
这是一个有趣的问题,我也为此编写了一个项目。
这里有些例子:
打印随机的BST。
BTPrinter.printRandomBST(100, 100);
38
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
28 82
/ \ / \
/ \ / \
/ \ / \
/ \ / \
5 31 / \
/ \ / \ / \
/ \ 30 36 / \
/ \ / / \ / \
/ \ 29 33 37 / \
/ \ / \ / \
/ \ 32 35 65 95
1 14 / / \ / \
/ \ / \ 34 / \ 94 97
0 2 / \ / \ / / \
\ 12 24 / \ 93 96 98
3 / \ / \ / \ / \
\ 9 13 16 25 / \ 84 99
4 / \ / \ \ / \ / \
7 10 15 23 26 59 74 83 86
/ \ \ / \ / \ / \ / \
6 8 11 22 27 56 60 73 76 85 91
/ / \ \ / / \ / \
20 / \ 61 67 75 79 88 92
/ \ 40 58 \ / \ / \ / \
18 21 / \ / 62 66 72 78 80 87 89
/ \ 39 54 57 \ / / \ \
17 19 / \ 64 69 77 81 90
50 55 / / \
/ \ 63 68 70
/ \ \
/ \ 71
47 53
/ \ /
/ \ 52
42 49 /
/ \ / 51
41 43 48
\
46
/
45
/
44
来自leetcode样式级别顺序数组的打印树,“#”表示路径终止符,其中下面没有节点。
BTPrinter.printTree("1,2,3,4,5,#,#,6,7,8,1,#,#,#,#,#,#,2,3,4,5,6,7,8,9,10,11,12,13,14,15");
1
/ \
2 3
/ \
/ \
4 5
/ \ / \
6 7 8 1
/ \
/ \
/ \
/ \
/ \
2 3
/ \ / \
/ \ / \
4 5 6 7
/ \ / \ / \ / \
8 9 10 11 12 13 14 15
我需要在我的一个项目中打印一个二叉树,因为我已经准备了一个java类TreePrinter
,示例输出之一是:
[+]
/ \
/ \
/ \
/ \
/ \
[*] \
/ \ [-]
[speed] [2] / \
[45] [12]
这是class TreePrinter
和class 的代码TextNode
。要打印任何树,您只需创建带有TextNode
类的等效树即可。
import java.util.ArrayList;
public class TreePrinter {
public TreePrinter(){
}
public static String TreeString(TextNode root){
ArrayList layers = new ArrayList();
ArrayList bottom = new ArrayList();
FillBottom(bottom, root); DrawEdges(root);
int height = GetHeight(root);
for(int i = 0; i s.length()) min = s.length();
if(!n.isEdge) s += "[";
s += n.text;
if(!n.isEdge) s += "]";
layers.set(n.depth, s);
}
StringBuilder sb = new StringBuilder();
for(int i = 0; i temp = new ArrayList();
for(int i = 0; i 0) temp.get(i-1).left = x;
temp.add(x);
}
temp.get(count-1).left = n.left;
n.left.depth = temp.get(count-1).depth+1;
n.left = temp.get(0);
DrawEdges(temp.get(count-1).left);
}
if(n.right != null){
int count = n.right.x - (n.x + n.text.length() + 2);
ArrayList temp = new ArrayList();
for(int i = 0; i 0) temp.get(i-1).right = x;
temp.add(x);
}
temp.get(count-1).right = n.right;
n.right.depth = temp.get(count-1).depth+1;
n.right = temp.get(0);
DrawEdges(temp.get(count-1).right);
}
}
private static void FillBottom(ArrayList bottom, TextNode n){
if(n == null) return;
FillBottom(bottom, n.left);
if(!bottom.isEmpty()){
int i = bottom.size()-1;
while(bottom.get(i).isEdge) i--;
TextNode last = bottom.get(i);
if(!n.isEdge) n.x = last.x + last.text.length() + 3;
}
bottom.add(n);
FillBottom(bottom, n.right);
}
private static boolean isLeaf(TextNode n){
return (n.left == null && n.right == null);
}
private static int GetHeight(TextNode n){
if(n == null) return 0;
int l = GetHeight(n.left);
int r = GetHeight(n.right);
return Math.max(l, r) + 1;
}
}
class TextNode {
public String text;
public TextNode parent, left, right;
public boolean isEdge;
public int x, depth;
public TextNode(String text){
this.text = text;
parent = null; left = null; right = null;
isEdge = false;
x = 0; depth = 0;
}
}
最后是用于打印给定样本的测试类:
public class Test {
public static void main(String[] args){
TextNode root = new TextNode("+");
root.left = new TextNode("*"); root.left.parent = root;
root.right = new TextNode("-"); root.right.parent = root;
root.left.left = new TextNode("speed"); root.left.left.parent = root.left;
root.left.right = new TextNode("2"); root.left.right.parent = root.left;
root.right.left = new TextNode("45"); root.right.left.parent = root.right;
root.right.right = new TextNode("12"); root.right.right.parent = root.right;
System.out.println(TreePrinter.TreeString(root));
}
}
您可以使用小程序非常容易地将其可视化。您需要打印以下项目。
将节点打印为半径可见的圆形
获取每个节点的坐标。
x坐标可以可视化为在按顺序遍历该节点之前访问该节点的节点数。
y坐标可以可视化为特定节点的深度。
打印父母与子女之间的界线
这可以通过将节点的x和y坐标以及每个节点的父级维护在单独的列表中来完成。
对于除根以外的每个节点,通过获取子节点和父节点的x和y坐标,将每个节点与其父节点连接。
private StringBuilder prettyPrint(Node root, int currentHeight, int totalHeight) {
StringBuilder sb = new StringBuilder();
int spaces = getSpaceCount(totalHeight-currentHeight + 1);
if(root == null) {
//create a 'spatial' block and return it
String row = String.format("%"+(2*spaces+1)+"s%n", "");
//now repeat this row space+1 times
String block = new String(new char[spaces+1]).replace("\0", row);
return new StringBuilder(block);
}
if(currentHeight==totalHeight) return new StringBuilder(root.data+"");
int slashes = getSlashCount(totalHeight-currentHeight +1);
sb.append(String.format("%"+(spaces+1)+"s%"+spaces+"s", root.data+"", ""));
sb.append("\n");
//now print / and \
// but make sure that left and right exists
char leftSlash = root.left == null? ' ':'/';
char rightSlash = root.right==null? ' ':'\\';
int spaceInBetween = 1;
for(int i=0, space = spaces-1; i<slashes; i++, space --, spaceInBetween+=2) {
for(int j=0; j<space; j++) sb.append(" ");
sb.append(leftSlash);
for(int j=0; j<spaceInBetween; j++) sb.append(" ");
sb.append(rightSlash+"");
for(int j=0; j<space; j++) sb.append(" ");
sb.append("\n");
}
//sb.append("\n");
//now get string representations of left and right subtrees
StringBuilder leftTree = prettyPrint(root.left, currentHeight+1, totalHeight);
StringBuilder rightTree = prettyPrint(root.right, currentHeight+1, totalHeight);
// now line by line print the trees side by side
Scanner leftScanner = new Scanner(leftTree.toString());
Scanner rightScanner = new Scanner(rightTree.toString());
// spaceInBetween+=1;
while(leftScanner.hasNextLine()) {
if(currentHeight==totalHeight-1) {
sb.append(String.format("%-2s %2s", leftScanner.nextLine(), rightScanner.nextLine()));
sb.append("\n");
spaceInBetween-=2;
}
else {
sb.append(leftScanner.nextLine());
sb.append(" ");
sb.append(rightScanner.nextLine()+"\n");
}
}
return sb;
}
private int getSpaceCount(int height) {
return (int) (3*Math.pow(2, height-2)-1);
}
private int getSlashCount(int height) {
if(height <= 3) return height -1;
return (int) (3*Math.pow(2, height-3)-1);
}
https://github.com/murtraja/java-binary-tree-printer
仅适用于1到2位数的整数(我很懒于使它通用)
这是水平视图的最简单解决方案。尝试了一堆例子。对于我的目的来说效果很好。更新自@ nitin-k的答案。
public void print(String prefix, BTNode n, boolean isLeft) {
if (n != null) {
print(prefix + " ", n.right, false);
System.out.println (prefix + ("|-- ") + n.data);
print(prefix + " ", n.left, true);
}
}
呼叫:
bst.print("", bst.root, false);
解:
|-- 80
|-- 70
|-- 60
|-- 50
|-- 40
|-- 30
|-- 20
|-- 10
node_length * nodes_count + space_length * spaces_count*
。GitHub上的代码: YoussefRaafatNasry / bst-ascii-visualization
07
/\
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
03 11
/\ /\
/ \ / \
/ \ / \
/ \ / \
/ \ / \
01 05 09 13
/\ /\ /\ /\
/ \ / \ / \ / \
00 02 04 06 08 10 12 14
visualize
函数,而是整个visualizer
类,包括头文件在内,大约200 loc。
对于那些寻找Rust解决方案的人:
pub struct Node {
pub value: i32,
left: Option<Box<Node>>,
right: Option<Box<Node>>
}
impl Node {
pub fn new(val: i32) -> Node {
Node {
value: val,
left: None,
right: None
}
}
pub fn getLeftNode(&self) -> Option<&Node> {
self.left.as_deref()
}
pub fn getRightNode(&self) -> Option<&Node> {
self.right.as_deref()
}
pub fn setLeftNode(&mut self, val: i32) -> &mut Node {
self.left = Some(Box::new(Node::new(val)));
self.left.as_deref_mut().unwrap()
}
pub fn setRightNode(&mut self, val: i32) -> &mut Node {
self.right = Some(Box::new(Node::new(val)));
self.right.as_deref_mut().unwrap()
}
fn visualizeTree(&self, level: u16, is_tail: bool, columns: &mut HashSet<u16>) {
let left = self.getLeftNode();
let right = self.getRightNode();
if right.is_some() {
right.unwrap().visualizeTree(level+1, false, columns);
}
if level > 0 {
for i in 0..level-1 {
if columns.contains(&i) {
print!("│ ");
} else {
print!(" ");
}
}
if is_tail {
println!("└── {}", self.value);
columns.remove(&(level-1));
columns.insert(level);
} else {
println!("┌── {}", self.value);
columns.insert(level);
columns.insert(level-1);
}
} else {
println!("{}", self.value);
}
if left.is_some() {
left.unwrap().visualizeTree(level+1, true, columns);
}
}
pub fn printTree(&self) {
let mut columns = HashSet::new();
columns.insert(0);
self.visualizeTree(0, true, &mut columns);
}
}
输出是这样的:
┌── 17
│ │ ┌── 3
│ │ │ └── 9
│ └── 2
│ └── 1
20
│ ┌── 7
│ │ │ ┌── 16
│ │ └── 15
└── 8
│ ┌── 11
└── 4
└── 13
在控制台中打印:
500
700 300
200 400
简单的代码:
public int getHeight()
{
if(rootNode == null) return -1;
return getHeight(rootNode);
}
private int getHeight(Node node)
{
if(node == null) return -1;
return Math.max(getHeight(node.left), getHeight(node.right)) + 1;
}
public void printBinaryTree(Node rootNode)
{
Queue<Node> rootsQueue = new LinkedList<Node>();
Queue<Node> levelQueue = new LinkedList<Node>();
levelQueue.add(rootNode);
int treeHeight = getHeight();
int firstNodeGap;
int internalNodeGap;
int copyinternalNodeGap;
while(true)
{
System.out.println("");
internalNodeGap = (int)(Math.pow(2, treeHeight + 1) -1);
copyinternalNodeGap = internalNodeGap;
firstNodeGap = internalNodeGap/2;
boolean levelFirstNode = true;
while(!levelQueue.isEmpty())
{
internalNodeGap = copyinternalNodeGap;
Node currNode = levelQueue.poll();
if(currNode != null)
{
if(levelFirstNode)
{
while(firstNodeGap > 0)
{
System.out.format("%s", " ");
firstNodeGap--;
}
levelFirstNode =false;
}
else
{
while(internalNodeGap>0)
{
internalNodeGap--;
System.out.format("%s", " ");
}
}
System.out.format("%3d",currNode.data);
rootsQueue.add(currNode);
}
}
--treeHeight;
while(!rootsQueue.isEmpty())
{
Node currNode = rootsQueue.poll();
if(currNode != null)
{
levelQueue.add(currNode.left);
levelQueue.add(currNode.right);
}
}
if(levelQueue.isEmpty()) break;
}
}
这是一台非常通用的树打印机。外观不是最好的,但是可以处理很多情况。如果可以弄清楚,请随时添加斜杠。
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,节点接口
package com.tomac120.NodePrinter;
/**
* Created by elijah on 6/28/16.
*/
public interface PrintableNode {
PrintableNodeInfo _getPrintableNodeInfo();
PrintableNode _getLeftChild();
PrintableNode _getRightChild();
}
一个由Vasya Novikov的答案改编而来的Scala解决方案,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)
}
}
另请参阅以下答案。
特别是,使用abego TreeLayout并不难默认设置生成如下所示的结果。
如果您尝试使用该工具,请注意以下警告:它按照添加顺序打印子项。对于一个BST,左与右很重要,我发现该库未经修改是不合适的。
此外,添加子项的方法仅将parent
and child
节点作为参数。(因此,要处理一堆节点,必须将第一个节点分开创建一个根。)
我最终在上面使用了此解决方案,对其进行了修改以使其接受类型<Node>
,以便可以访问Node
的左(和右)(子代)。
这是可视化树的另一种方法:将节点另存为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){
}
}
}
这是测试它的代码:
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();
输出看起来像这样:
using map...
{
Map<Integer,String> m = new LinkedHashMap<>();
tn.printNodeWithLvl(node,l,m);
for(Entry<Integer, String> map :m.entrySet()) {
System.out.println(map.getValue());
}
then....method
private void printNodeWithLvl(Node node,int l,Map<Integer,String> m) {
if(node==null) {
return;
}
if(m.containsKey(l)) {
m.put(l, new StringBuilder(m.get(l)).append(node.value).toString());
}else {
m.put(l, node.value+"");
}
l++;
printNodeWithLvl( node.left,l,m);
printNodeWithLvl(node.right,l,m);
}
}
这是我可以实现的最简单的版本之一。希望对您有帮助
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def add(self, data):
if data < self.data:
if self.left is None:
self.left = Node(data)
else:
self.left.add(data)
if data > self.data:
if self.right is None:
self.right = Node(data)
else:
self.right.add(data)
def display(self):
diff = 16
start = 50
c = ' '
this_level = [(self, start)]
while this_level:
next_level = list()
last_line = ''
for node, d in this_level:
line = last_line + c*(d - len(last_line)) + str(node.data)
print(line, end='\r')
last_line = line
if node.left:
next_level.append((node.left, d - diff))
if node.right:
next_level.append((node.right, d + diff))
this_level = next_level
diff = max(diff//2, 2)
print('\n')
if __name__ == '__main__':
from random import randint, choice
values = [randint(0, 100) for _ in range(10)]
bst = Node(choice(values))
for data in values:
bst.add(data)
bst.display()