[2019.01.14] BST Level-Order Traversal

Problem

Objective 
Today, we're going further with Binary Search Trees. Check out the Tutorial tab for learning materials and an instructional video!

Task 
A level-order traversal, also known as a breadth-first search, visits each level of a tree's nodes from left to right, top to bottom. You are given a pointer, , pointing to the root of a binary search tree. Complete the levelOrder function provided in your editor so that it prints the level-order traversal of the binary search tree.

Hint: You'll find a queue helpful in completing this challenge.

Input Format

The locked stub code in your editor reads the following inputs and assembles them into a BST: 
The first line contains an integer,  (the number of test cases). 
The  subsequent lines each contain an integer, , denoting the value of an element that must be added to the BST.

Output Format

Print the  value of each node in the tree's level-order traversal as a single line of  space-separated integers.

Sample Input

6
3
5
4
7
2
1

Sample Output

3 2 5 1 4 7 

Explanation

The input forms the following binary search tree: 

We traverse each level of the tree from the root downward, and we process the nodes at each level from left to right. The resulting level-order traversal is , and we print these data values as a single line of space-separated integers.

How I solved the problem

# Binary Search Tree와 Queue를 함께 써서 풀어보는 문제..!
# Node와 Binary Search Tree의 개념을 한 번 더 되짚어보기

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67

import java.util.*; import java.io.*; class Node{     Node left,right;     int data;     Node(int data){         this.data=data;         left=right=null;     } } class Solution{   static void levelOrder(Node root){ //binary search tree와 queue를 혼합한 문제       //Write your code here       Queue<Node> queue = new LinkedList();       //Queue이기 때문에 first in, first out의 개념이 들어가 있음       queue.add(root);              String result = "";         while (!queue.isEmpty()) {           Node current = queue.remove();           //먼저 현재 current node의 데이터를 result에 넣음           result += " " + current.data;                      //왼쪽을 먼저 읽기 때문에 왼쪽이 있는지 확인한 후 add           if (current.left != null) {               queue.add(current.left);           }                      //다음에 오른쪽이 있는지 확인한 후 add           if (current.right != null) {               queue.add(current.right);           }       }         System.out.println(result.substring(1)); }   public static Node insert(Node root,int data){         if(root==null){             return new Node(data);         }         else{             Node cur;             if(data<=root.data){                 cur=insert(root.left,data);                 root.left=cur;             }             else{                 cur=insert(root.right,data);                 root.right=cur;             }             return root;         }     }     public static void main(String args[]){             Scanner sc=new Scanner(System.in);             int T=sc.nextInt();             Node root=null;             while(T-->0){                 int data=sc.nextInt();                 root=insert(root,data);             }             levelOrder(root);         }     } Colored by Color Scripter

cs

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