我们得到了一个需要编写代码的作业：

• 二叉搜索树
• 树必须是完整的，而不是完美的（这意味着所有不在最低级别或第二低级别的节点都应有2个子节点，而最低级别的节点应尽可能远）
• 我们需要按级别顺序插入树中

• 因此，如果我有一个包含元素{0, 1, 2, 3, 4, 5, 6, 7}的数组，root应该是4，2, 1, 3, 0在左侧和6, 5, 7右侧。

• 级别订单插入为： 4, 2, 6, 1, 3, 5, 7, 0

• 仅将Array的中间部分放在根目录是行不通的。如果有1到9个元素的数组，则将有4个作为根（java中的int值，double为4.5），右侧将有5个元素，左侧有4个元素。这不是一棵完整的树。甚至不是完美的树。

我的代码只能向左或向右插入，具体取决于它是否
大于或小于根，而不插入级别顺序。该Anytype x
参数是要插入的价值，同时BinaryNode t为当前
节点，我们在树（这就是我们的比较，如果我们需要新的插入
值向左或向右）

private BinaryNode<AnyType> insert( AnyType x, BinaryNode<AnyType> t )
{
    if( t == null )
        return new BinaryNode<>( x, null, null );

    int compareResult = x.compareTo( t.element );

    if( compareResult < 0 )
        t.left = insert( x, t.left );
    else if( compareResult > 0 )
        t.right = insert( x, t.right );
    else
        ;  // Duplicate; do nothing
    return t;
}

如何按级别顺序插入并仍然维护二进制搜索树？
我应该使用某种形式的递归吗？

我的整个程序

import java.nio.BufferUnderflowException;
import java.util.*;

import static java.lang.Math.pow;

/**
 * Implements an unbalanced binary search tree.
 * Note that all "matching" is based on the compareTo method.
 * @author Mark Allen Weiss
 */
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>>
{
    /**
     * Construct the tree.
     */
    public BinarySearchTree( )
    {
        root = null;
    }

    /**
     * Insert into the tree; duplicates are ignored.
     * @param x the item to insert.
     */
    public void insert( AnyType x )
    {
        root = insert( x, root );
    }

    /**
     * Test if the tree is logically empty.
     * @return true if empty, false otherwise.
     */
    public boolean isEmpty( )
    {
        return root == null;
    }

    /**
     * Print the tree contents in sorted order.
     */
    public void printTree( )
    {
        if( isEmpty( ) )
            System.out.println( "Empty tree" );
        else
            printTree( root );
    }

    /**
     * Internal method to insert into a subtree.
     * @param x the item to insert.
     * @param t the node that roots the subtree.
     * @return the new root of the subtree.
     */
    private BinaryNode<AnyType> insert( AnyType x, BinaryNode<AnyType> t )
    {
        if( t == null )
            return new BinaryNode<>( x, null, null );

        int compareResult = x.compareTo( t.element );

        if( compareResult < 0 )
            t.left = insert( x, t.left );
        else if( compareResult > 0 )
            t.right = insert( x, t.right );
        else
            ;  // Duplicate; do nothing
        return t;
    }

    /**
     * Internal method to print a subtree in sorted order.
     * @param t the node that roots the subtree.
     */
    private void printTree( BinaryNode<AnyType> t )
    {
        if( t != null )
        {
            printTree( t.left );
            System.out.println( t.element );
            printTree( t.right );
        }
    }

    /**
     * Internal method to compute height of a subtree.
     * @param t the node that roots the subtree.
     */
    private int height( BinaryNode<AnyType> t )
    {
        if( t == null )
            return -1;
        else
            return 1 + Math.max( height( t.left ), height( t.right ) );    
    }

    // Basic node stored in unbalanced binary search trees
    private static class BinaryNode<AnyType>
    {
            // Constructors
        BinaryNode( AnyType theElement )
        {
            this( theElement, null, null );
        }

        BinaryNode( AnyType theElement, BinaryNode<AnyType> lt, BinaryNode<AnyType> rt )
        {
            element  = theElement;
            left     = lt;
            right    = rt;
        }

        AnyType element;            // The data in the node
        BinaryNode<AnyType> left;   // Left child
        BinaryNode<AnyType> right;  // Right child
    }

      /** The tree root. */
    private BinaryNode<AnyType> root;

        // Test program
    public static void main( String [ ] args )
    {
        BinarySearchTree<Integer> t = new BinarySearchTree<>( );

        t.insert(2);
        t.insert(1);
        t.insert(3);
        t.printTree();
    }
}