[BOJ] 1991번: 트리 순회

출처 : https://www.acmicpc.net/problem/1991

1991번: 트리 순회

 

 

node 클래스를 만들어서 left 자식과 right 자식을 관리해주면 된다. 쉬운 문제였다 !

package com.ll.boj.silver.p1991;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {
    static class node {
        int left;
        int right;

        public node(int left, int right) {
            this.left = left;
            this.right = right;
        }
    }
    static List<node>[] graph;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        sc.nextLine();

        graph = new List[n];
        for(int i=0; i<n; i++) graph[i] = new ArrayList<>();


        for(int i=0; i<n; i++) {
            String[] tmp = sc.nextLine().split(" ");
            int root = tmp[0].charAt(0) - 'A';
            int left = tmp[1].charAt(0) - 'A';
            int right = tmp[2].charAt(0) - 'A';
            graph[root].add(new node(left, right));

        }
        preorder(0);
        System.out.println();

        inorder(0);
        System.out.println();

        postorder(0);
    }

    private static void postorder(int start) {
        for(node x : graph[start]) {
            if (x.left != -19) postorder(x.left);
            if (x.right != -19) postorder(x.right);
            System.out.print((char)(start + 'A'));
        }
    }

    private static void inorder(int start) {
        for(node x : graph[start]) {
            if (x.left != -19) inorder(x.left);
            System.out.print((char)(start + 'A'));
            if (x.right != -19) inorder(x.right);
        }
    }

    private static void preorder(int start) {
        for(node x : graph[start]) {
            System.out.print((char)(start + 'A'));
            if (x.left != -19) preorder(x.left);
            if (x.right != -19) preorder(x.right);
        }
    }
}