挑战程序设计竞赛（算法和数据结构）——8.2有根树的表达的JAVA实现

题目：

思路：

需要特别主义的是，建立Node数组对象后，必须先对Node数组的每个元素分配内存/初始化，否则会报错，这个bug折磨了我一上午：

Exception in thread “main” java.lang.NullPointerException: Cannot assign field “parent” because “T[c[j]]” is null
at TreeDemo.main

也就是务必要注意这个部分的代码：

        for(int i=0;i 
完整代码如下：
 
import java.io.BufferedInputStream;
import java.util.Scanner;

public class TreeDemo {
    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[] 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：
13
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 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0
 
输出：
 
node 0: parent = -1, depth = 0, root, [1 4 10 ]
node 1: parent = 0, depth = 1, internal node, [2 3 ]
node 2: parent = 1, depth = 2, internal node, []
node 3: parent = 1, depth = 2, internal node, []
node 4: parent = 0, depth = 1, internal node, [5 6 7 ]
node 5: parent = 4, depth = 2, internal node, []
node 6: parent = 4, depth = 2, internal node, []
node 7: parent = 4, depth = 2, internal node, [8 9 ]
node 8: parent = 7, depth = 3, internal node, []
node 9: parent = 7, depth = 3, internal node, []
node 10: parent = 0, depth = 1, internal node, [11 12 ]
node 11: parent = 10, depth = 2, internal node, []
node 12: parent = 10, depth = 2, internal node, []