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
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);
}
}
其他回答
我已经创建了简单的二叉树打印机。您可以随心所欲地使用和修改它,但无论如何它都没有优化。我认为这里有很多东西可以改进;)
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。
输出的例子:
z
├── c
│ ├── a
│ └── b
├── d
├── e
│ └── asdf
└── f
代码:
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 + " ");
}
}
}
}
附注:这个答案并不完全关注“二叉”树——相反,它打印了各种类型的树。解决方案的灵感来自linux中的“树”命令。
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。它使用行自上而下地打印。
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;
}
}
}
要在树中使用它,让Node类实现PrintableNode。
示例输出:
2952:0
┌───────────────────────┴───────────────────────┐
1249:-1 5866:0
┌───────────┴───────────┐ ┌───────────┴───────────┐
491:-1 1572:0 4786:1 6190:0
┌─────┘ └─────┐ ┌─────┴─────┐
339:0 5717:0 6061:0 6271:0
这是水平视图最简单的解决方案。我举了很多例子。很适合我的目的。更新自@ ntin -k的回答。
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);
}
}
电话:
bst.print("", bst.root, false);
解决方案:
|-- 80
|-- 70
|-- 60
|-- 50
|-- 40
|-- 30
|-- 20
|-- 10
根据VasyaNovikov的回答。改进了一些Java魔术:泛型和函数接口。
/**
* 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);
将输出如下内容
└── rootNode
├── childNode1
├── childNode2
│ ├── childNode2.1
│ ├── childNode2.2
│ └── childNode2.3
├── childNode3
└── childNode4