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。

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

你的树每一层需要两倍的距离:

       a
      / \
     /   \
    /     \
   /       \
   b       c
  / \     / \
 /   \   /   \
 d   e   f   g
/ \ / \ / \ / \
h i j k l m n o

你可以将你的树保存在一个数组的数组中，每个数组对应一个深度:

[[a],[b,c],[d,e,f,g],[h,i,j,k,l,m,n,o]]

如果你的树没有满，你需要在数组中包含空值:

       a
      / \
     /   \
    /     \
   /       \
   b       c
  / \     / \
 /   \   /   \
 d   e   f   g
/ \   \ / \   \
h i   k l m   o
[[a],[b,c],[d,e,f,g],[h,i, ,k,l,m, ,o]]

然后你可以遍历数组来打印你的树，根据深度打印第一个元素之前和元素之间的空格，根据下一层数组中对应的元素是否被填充打印行。 如果您的值可以超过一个字符长，您需要在创建数组表示时找到最长的值，并相应地乘以所有宽度和行数。

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

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