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
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
在控制台打印:
500
700 300
200 400
简单代码:
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;
}
}
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。
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
对于输入
8 4 12 2 6 10 14 1 3 5 7 9 11 13 20 15
这是@anurag回答的一个变体——看到额外的|让我很烦