Create a tree in level order - GeeksforGeeks

08 Feb, 2022
Given an array of elements, the task is to insert these elements in level order and construct a tree.
Input : arr[] = {10, 20, 30, 40, 50, 60}
Output :         10
              /      \
             20       30
            /  \     /
          40    50  60
The task is to construct a whole tree from a given array. To insert in level order in an already constructed tree, please see Insertion in a Binary Tree in level orderThe task is to store data in a binary tree but in level order.To do so, we will proceed as follows: 1. Whenever a new Node is added to the binary tree, the address of the node is pushed into a queue. 2. A Node address will stay in the queue until both its children’s Nodes do not get filled. 3. Once both the children’s Nodes get filled up, the parent Node is popped from the queue.Here is the code to perform the above mentioned data entry.
C++
Java
Python3
C#
Javascript
// CPP program to construct a binary tree in level order.#include <bits/stdc++.h>using namespace std; struct Node {    int key;    Node* left;    Node* right;}; // Function to create a node with 'value' as the data// stored in it.// Both the children of this new Node are initially null.struct Node* newNode(int value){    Node* n = new Node;    n->key = value;    n->left = NULL;    n->right = NULL;    return n;} struct Node* insertValue(struct Node* root, int value,                         queue<Node *>& q){    Node* node = newNode(value);    if (root == NULL)        root = node;     // The left child of the current Node is    // used if it is available.    else if (q.front()->left == NULL)        q.front()->left = node;     // The right child of the current Node is used    // if it is available. Since the left child of this    // node has already been used, the Node is popped    // from the queue after using its right child.    else {        q.front()->right = node;        q.pop();    }     // Whenever a new Node is added to the tree, its    // address is pushed into the queue.    // So that its children Nodes can be used later.    q.push(node);    return root;} // This function mainly calls insertValue() for// all array elements. All calls use same queue.Node* createTree(int arr[], int n){    Node* root = NULL;    queue<Node*> q;    for (int i = 0; i < n; i++)      root = insertValue(root, arr[i], q);    return root;} // This is used to verify the logic.void levelOrder(struct Node* root){    if (root == NULL)        return;    queue<Node*> n;    n.push(root);    while (!n.empty()) {        cout << n.front()->key << " ";        if (n.front()->left != NULL)            n.push(n.front()->left);        if (n.front()->right != NULL)            n.push(n.front()->right);        n.pop();    }} // Driver codeint main(){    int arr[] = { 10, 20, 30, 40, 50, 60 };    int n = sizeof(arr) / sizeof(arr[0]);    Node* root = createTree(arr, n);    levelOrder(root);    return 0;}
// JAVA program to construct a// binary tree in level// order.import java.util.*;class GFG{ static class Node{  int key;  Node left;  Node right;};   static Node root = null; static Queue<Node> q =       new LinkedList<>();   // Function to create a node// with 'value' as the data// stored in it.// Both the children of this// new Node are initially null.static Node newNode(int value){  Node n = new Node();  n.key = value;  n.left = null;  n.right = null;  return n;} static void insertValue(int value){  Node node = newNode(value);  if (root == null)    root = node;   // The left child of the  // current Node is used  // if it is available.  else if (q.peek().left == null)    q.peek().left = node;   // The right child of the current  // Node is used if it is available.  // Since the left child of this  // node has already been used, the  // Node is popped from the queue  // after using its right child.  else  {    q.peek().right = node;    q.remove();  }   // Whenever a new Node is added  // to the tree, its address is  // pushed into the queue. So that  // its children Nodes can be used  // later.  q.add(node); } // This function mainly calls// insertValue() for all array// elements. All calls use same// queue.static void createTree(int arr[],                       int n){  for (int i = 0; i < n; i++)    insertValue(arr[i]);} // This is used to verify// the logic.static void levelOrder(Node root){  if (root == null)    return;  Queue<Node> n =        new LinkedList<>();  n.add(root);  while (!n.isEmpty())  {    System.out.print(n.peek().key +                     " ");    if (n.peek().left != null)      n.add(n.peek().left);    if (n.peek().right != null)      n.add(n.peek().right);    n.remove();  }} // Driver codepublic static void main(String[] args){  int arr[] = {10, 20, 30,               40, 50, 60};  int n = arr.length;  createTree(arr, n);  levelOrder(root);}} // This code is contributed by Rajput-Ji
# Python3 program to construct# a binary tree in level order. # Importing Queue for use in# Level Order Traversalfrom collections import deque # Node class for holding the Binary Treeclass node:    def __init__(self, data = None):        self.data = data        self.left = None        self.right = None Q = deque() # Helper function helps us in adding data# to the tree in Level Orderdef insertValue(data, root):    newnode = node(data)    if Q:        temp = Q[0]    if root == None:        root = newnode             # The left child of the current Node is    # used if it is available.    elif temp.left == None:        temp.left = newnode         # The right child of the current Node is used    # if it is available. Since the left child of this    # node has already been used, the Node is popped    # from the queue after using its right child.    elif temp.right == None:        temp.right = newnode        atemp = Q.popleft()         # Whenever a new Node is added to the tree,    # its address is pushed into the queue.    # So that its children Nodes can be used later.    Q.append(newnode)    return root # Function which calls add which is responsible# for adding elements one by onedef createTree(a, root):    for i in range(len(a)):        root = insertValue(a[i], root)    return root # Function for printing level order traversaldef levelOrder(root):    Q = deque()    Q.append(root)    while Q:        temp = Q.popleft()        print(temp.data, end = ' ')        if temp.left != None:            Q.append(temp.left)        if temp.right != None:            Q.append(temp.right) # Driver Codea = [ 10, 20, 30, 40, 50, 60 ]root = Noneroot = createTree(a, root) levelOrder(root) # This code is contributed by code_freak
// C# program to construct a// binary tree in level// order.using System;using System.Collections.Generic;class GFG {         class Node    {        public int key;        public Node left, right;    };         static Node root = null;         static List<Node> q = new List<Node>();            // Function to create a node    // with 'value' as the data    // stored in it.    // Both the children of this    // new Node are initially null.    static Node newNode(int value)    {      Node n = new Node();      n.key = value;      n.left = null;      n.right = null;      return n;    }          static void insertValue(int value)    {      Node node = newNode(value);      if (root == null)        root = node;            // The left child of the      // current Node is used      // if it is available.      else if (q[0].left == null)        q[0].left = node;            // The right child of the current      // Node is used if it is available.      // Since the left child of this      // node has already been used, the      // Node is popped from the queue      // after using its right child.      else      {        q[0].right = node;        q.RemoveAt(0);      }            // Whenever a new Node is added      // to the tree, its address is      // pushed into the queue. So that      // its children Nodes can be used      // later.      q.Add(node);          }          // This function mainly calls    // insertValue() for all array    // elements. All calls use same    // queue.    static void createTree(int[] arr, int n)    {      for (int i = 0; i < n; i++)        insertValue(arr[i]);    }          // This is used to verify    // the logic.    static void levelOrder(Node root)    {      if (root == null)        return;               List<Node> n = new List<Node>();      n.Add(root);      while (n.Count > 0)      {        Console.Write(n[0].key + " ");        if (n[0].left != null)          n.Add(n[0].left);        if (n[0].right != null)          n.Add(n[0].right);        n.RemoveAt(0);      }    }   // Driver code  static void Main() {    int[] arr = {10, 20, 30, 40, 50, 60};    int n = arr.Length;    createTree(arr, n);    levelOrder(root);  }} // This code is contributed by divyeshrabadiya07.
<script> // Javascript program to construct a// binary tree in level// order. // Binary Tree nodeclass Node{    constructor(value)    {        this.left = null;        this.right = null;        this.key = value;    }} let root = null;let q = []; // Function to create a node// with 'value' as the data// stored in it.// Both the children of this// new Node are initially null.function newNode(value){    let n = new Node(value);    return n;} function insertValue(value){    let node = newNode(value);    if (root == null)        root = node;         // The left child of the    // current Node is used    // if it is available.    else if (q[0].left == null)        q[0].left = node;         // The right child of the current    // Node is used if it is available.    // Since the left child of this    // node has already been used, the    // Node is popped from the queue    // after using its right child.    else    {        q[0].right = node;        q.shift();    }         // Whenever a new Node is added    // to the tree, its address is    // pushed into the queue. So that    // its children Nodes can be used    // later.    q.push(node); } // This function mainly calls// insertValue() for all array// elements. All calls use same// queue.function createTree(arr, n){    for(let i = 0; i < n; i++)        insertValue(arr[i]);} // This is used to verify// the logic.function levelOrder(root){    if (root == null)        return;             let n = [];    n.push(root);         while (n.length > 0)    {        document.write(n[0].key + " ");        if (n[0].left != null)            n.push(n[0].left);        if (n[0].right != null)            n.push(n[0].right);                     n.shift();    }} // Driver codelet arr = [ 10, 20, 30, 40, 50, 60 ];let n = arr.length; createTree(arr, n);levelOrder(root); // This code is contributed by suresh07 </script>
10 20 30 40 50 60
 
Time Complexity: O(n)
code_freak
Rajput-Ji
mohamedshamir
suresh07
divyeshrabadiya07
simmytarika5
cpp-queue
tree-level-order
Tree
Tree
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