我一直在尝试创建一个递归方法,它将创建一个完整的二叉搜索树。此方法返回对此树的根的引用。作为参数,我传递树的深度以及存储在当前子树的根中的数字。我已经设法为2个基本情况设计解决方案,当深度为0和1时,但是当我尝试一个大于1的数字时,我只得到0级和1级实例化而不是下一个。任何帮助都会很棒
public class BinaryNode {
private int data;
private BinaryNode left, right;
public BinaryNode(int initialData, BinaryNode initialLeft,
BinaryNode initialRight) {
data = initialData;
left = initialLeft;
right = initialRight;
}
public static BinaryNode BSTFactory(int top,int depth) {
BinaryNode root=new BinaryNode(top,null,null);
BinaryNode leftChild,rightChild;
if(depth==0)
//return root;
if(depth==1){
//create 2 children left and right
leftChild=new BinaryNode(top-1,null,null);
rightChild=new BinaryNode(top+1,null,null);
root=new BinaryNode(top,leftChild,rightChild);
//return root;
}
if(depth>1){
leftChild=BSTFactory(top-1,depth-1);
rightChild=BSTFactory(top+1,depth-1);
root=new BinaryNode(top,leftChild,rightChild);
//return root;
}
return root;
}
public class Applications {
public static void main(String[] args){
BinaryNode root2=BinaryNode.BSTFactory(8, 2);
System.out.println(root2.toString());
}
}
This is the output:
data: 8
8's left: data: 7
7's left: null
7's right: null
8's right: data: 9
9's left: null
9's right: null
答案 0 :(得分:1)
当空树由null
表示时,通常不需要多个基本案例。
public class BinaryNode {
public static BinaryNode bstFactory( int data, int depth ) {
if ( depth >= 31 )
throw new IllegalArgumentException( "too deep for integer data" );
else if ( depth < 0 )
throw new IllegalArgumentException( "non-sensical depth" );
return ( depth == 0 )
? null
: new BinaryNode(
data,
bstFactory( data - ( 1 << depth ), depth - 1 ),
bstFactory( data + ( 1 << depth ), depth - 1 )
);
}
BinaryNode( int data, BinaryNode left, BinaryNode right ) { /*...*/ }
}