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

我需要在我的一个项目中打印一个二叉树,为此我准备了一个java类TreePrinter,其中一个示例输出是:

                [+]
               /   \
              /     \
             /       \
            /         \
           /           \
        [*]             \
       /   \             [-]
[speed]     [2]         /   \
                    [45]     [12]

下面是TreePrinter类和TextNode类的代码。为了打印任何树,你可以用TextNode类创建一个等效的树。


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,这里是一个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));
    }
}

其他回答

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。

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;
        }

    }

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。

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
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