题目：
有了有根树这篇博文的代码基础，二叉树很快就能写出来，只要把左右节点指针进行改写即可！
https://blog.csdn.net/weixin_42887138/article/details/121472382
import java.io.BufferedInputStream;
import java.util.Scanner;
public class BinaryTree {
public static class Node{
int parent, left, right;
Node(int parent, int left, int right){
this.parent = parent;
this.left = left;
this.right = right;
}
}
public static void main(String[] args) {
Scanner cin = new Scanner(new BufferedInputStream(System.in));
System.out.println("请输入树的节点数：");
int n = cin.nextInt();
Node[] T = new Node[10000];
int[] H = new int[10000];
int[] D = new int[10000];
System.out.println("接下来，每行输入节点的编号id，度k，以及第1到第k个子节点的编号c1,c2...ck:");
for(int i=0;i
输入：
请输入树的节点数：
9
接下来，每行输入节点的编号id，度k，以及第1到第k个子节点的编号c1,c2...ck:
0 1 4
1 2 3
2 -1 -1
3 -1 -1
4 5 8
5 6 7
6 -1 -1
7 -1 -1
8 -1 -1
输出：
node 0: parent = -1, sibling = -1, degree = 2, depth = 0, height = 3, root
node 1: parent = 0, sibling = 4, degree = 2, depth = 1, height = 1, internal node
node 2: parent = 1, sibling = 3, degree = 0, depth = 2, height = 0, leaf
node 3: parent = 1, sibling = 2, degree = 0, depth = 2, height = 0, leaf
node 4: parent = 0, sibling = 1, degree = 2, depth = 1, height = 2, internal node
node 5: parent = 4, sibling = 8, degree = 2, depth = 2, height = 1, internal node
node 6: parent = 5, sibling = 7, degree = 0, depth = 3, height = 0, leaf
node 7: parent = 5, sibling = 6, degree = 0, depth = 3, height = 0, leaf
node 8: parent = 4, sibling = 5, degree = 0, depth = 2, height = 0, leaf
