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

Posted 2021-12-18 小乖乖的臭坏坏

题目：

思路：

注意

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<n;i++) 
            T[i] = new Node(-1, -1, -1);//树节点的初始化
        

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<n;i++) 
            T[i] = new Node(-1, -1, -1);//树节点的初始化
        
        //建立树型结构，定义节点联系
        for(int i=0;i<n;i++)
            int id = cin.nextInt();//输入节点编号
            int k = cin.nextInt();//输入度
            int[] c = new int[k];//初始化子节点编号数组
            for(int j=0;j<k;j++)
                c[j] = cin.nextInt();
                T[c[j]].parent = id;//树里面的每一个子节点的父节点都对应id
                if(j==0)T[id].left = c[j];//如果是第一个子节点，则赋值给父节点的左节点
                if(j!=0)T[c[j-1]].right = c[j];//如果不是第一个子节点，赋值给上一个节点的右节点
            
        

        setDepth(T, D, 0,0);

        for (int i=0;i<n;i++)
            System.out.print("node " + i + ": parent = " + T[i].parent + ", depth = " + getDepth(T, i));
            printChildren(T, i);
            System.out.println();
        
    

    public static void setDepth(Node[] T, int[] D, int u, int p)
        D[u] = p;
        if(T[u].right != -1)
            setDepth(T, D , T[u].right, p);
        
        if(T[u].left != -1)
            setDepth(T, D, T[u].left, p+1);
        
    

    public static int getDepth(Node[] T, int u)
        int d = 0;
        while(T[u].parent != -1)
            u = T[u].parent;
            d++;
        
        return d;
    

    public static void printChildren(Node[] T, int u)

        if(T[u].left == -1)
            System.out.print(", internal node, []");
        
        else//只要不是叶子节点，都应该有子节点的打印
            if(T[u].parent == -1)
                System.out.print(", root, ");
            //没有父节点，即自己是根节点
            else if(T[u].parent!=-1 && T[u].left!=-1)
                System.out.print(", internal node, ");
            //没有父节点和子节点
            int c = T[u].left;
            System.out.print("[");
            while (c != -1)
                System.out.print(c + " ");
                c = T[c].right;
            
            System.out.print("]");
        
    

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, []