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Construct BST from its given level order traversal

  • Difficulty Level : Hard
  • Last Updated : 07 Jul, 2021

Construct the BST (Binary Search Tree) from its given level order traversal.
Examples: 
 

Input : arr[] = {7, 4, 12, 3, 6, 8, 1, 5, 10}
Output : BST: 
        7        
       / \       
      4   12      
     / \  /     
    3  6 8    
   /  /   \
  1   5   10

 

The idea is to use the Recursion:- 
We know that the first element will always be the root of tree and second element will be the left child and third element will be the right child (if fall in the range), and so on for all the remaining elements.
1) First pick the first element of the array and make it root. 
2) Pick the second element, if it’s value is smaller than root node value make it left child, 
3) Else make it right child 
4) Now recursively call the step (2) and step (3) to make a BST from its level Order Traversal.
Below is the implementation of above approach: 
 

C++




// C++ implementation to construct a BST
// from its level order traversal
#include <bits/stdc++.h>
 
using namespace std;
 
 
// node of a BST
struct Node
{
    int data;
    Node *left, *right;
};
 
 
// function to get a new node
Node* getNode(int data)
{
    // Allocate memory
    Node *newNode =
        (Node*)malloc(sizeof(Node));
     
    // put in the data   
    newNode->data = data;
    newNode->left = newNode->right = NULL;   
    return newNode;
}
 
 
// function to construct a BST from
// its level order traversal
Node *LevelOrder(Node *root , int data)
{
     if(root==NULL){   
        root = getNode(data);
        return root;
     }
     if(data <= root->data)
     root->left = LevelOrder(root->left, data);
     else
     root->right = LevelOrder(root->right, data);
     return root;    
}
 
Node* constructBst(int arr[], int n)
{
    if(n==0)return NULL;
    Node *root =NULL;
 
    for(int i=0;i<n;i++)
    root = LevelOrder(root , arr[i]);
     
    return root;
}
 
// function to print the inorder traversal
void inorderTraversal(Node* root)
{
    if (!root)
        return;
     
    inorderTraversal(root->left);
    cout << root->data << " ";
    inorderTraversal(root->right);   
}
 
 
// Driver program to test above
int main()
{
    int arr[] = {7, 4, 12, 3, 6, 8, 1, 5, 10};
    int n = sizeof(arr) / sizeof(arr[0]);
     
    Node *root = constructBst(arr, n);
     
    cout << "Inorder Traversal: ";
    inorderTraversal(root);
    return 0;   
}

Java




// Java implementation to construct a BST
// from its level order traversal
class GFG
{
 
// node of a BST
static class Node
{
    int data;
    Node left, right;
};
 
 
// function to get a new node
static Node getNode(int data)
{
    // Allocate memory
    Node newNode = new Node();
     
    // put in the data
    newNode.data = data;
    newNode.left = newNode.right = null;
    return newNode;
}
 
 
// function to construct a BST from
// its level order traversal
static Node LevelOrder(Node root , int data)
{
    if(root == null)
    {
        root = getNode(data);
        return root;
    }
    if(data <= root.data)
    root.left = LevelOrder(root.left, data);
    else
    root.right = LevelOrder(root.right, data);
    return root;    
}
 
static Node constructBst(int arr[], int n)
{
    if(n == 0)return null;
    Node root = null;
 
    for(int i = 0; i < n; i++)
    root = LevelOrder(root , arr[i]);
     
    return root;
}
 
// function to print the inorder traversal
static void inorderTraversal(Node root)
{
    if (root == null)
        return;
     
    inorderTraversal(root.left);
    System.out.print( root.data + " ");
    inorderTraversal(root.right);
}
 
 
// Driver code
public static void main(String args[])
{
    int arr[] = {7, 4, 12, 3, 6, 8, 1, 5, 10};
    int n = arr.length;
     
    Node root = constructBst(arr, n);
     
    System.out.print( "Inorder Traversal: ");
    inorderTraversal(root);
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python implementation to construct a BST
# from its level order traversal
import math
 
# node of a BST
class Node:
    def __init__(self,data):
        self.data = data
        self.left = None
        self.right = None
 
# function to get a new node
def getNode( data):
     
    # Allocate memory
    newNode = Node(data)
     
    # put in the data
    newNode.data = data
    newNode.left =None
    newNode.right = None
    return newNode
 
# function to construct a BST from
# its level order traversal
def LevelOrder(root , data):
    if(root == None):
        root = getNode(data)
        return root
     
    if(data <= root.data):
        root.left = LevelOrder(root.left, data)
    else:
        root.right = LevelOrder(root.right, data)
    return root    
 
def constructBst(arr, n):
    if(n == 0):
        return None
    root = None
 
    for i in range(0, n):
        root = LevelOrder(root , arr[i])
     
    return root
 
# function to print the inorder traversal
def inorderTraversal( root):
    if (root == None):
        return None
     
    inorderTraversal(root.left)
    print(root.data,end = " ")
    inorderTraversal(root.right)
 
# Driver program
if __name__=='__main__':
 
    arr = [7, 4, 12, 3, 6, 8, 1, 5, 10]
    n = len(arr)
     
    root = constructBst(arr, n)
     
    print("Inorder Traversal: ", end = "")
    root = inorderTraversal(root)
         
# This code is contributed by Srathore

C#




// C# implementation to construct a BST
// from its level order traversal
using System;
 
class GFG
{
 
// node of a BST
public class Node
{
    public int data;
    public Node left, right;
};
 
// function to get a new node
static Node getNode(int data)
{
    // Allocate memory
    Node newNode = new Node();
     
    // put in the data
    newNode.data = data;
    newNode.left = newNode.right = null;
    return newNode;
}
 
// function to construct a BST from
// its level order traversal
static Node LevelOrder(Node root,
                       int data)
{
    if(root == null)
    {
        root = getNode(data);
        return root;
    }
     
    if(data <= root.data)
        root.left = LevelOrder(root.left, data);
    else
        root.right = LevelOrder(root.right, data);
    return root;    
}
 
static Node constructBst(int []arr, int n)
{
    if(n == 0) return null;
    Node root = null;
 
    for(int i = 0; i < n; i++)
    root = LevelOrder(root, arr[i]);
     
    return root;
}
 
// function to print the inorder traversal
static void inorderTraversal(Node root)
{
    if (root == null)
        return;
     
    inorderTraversal(root.left);
    Console.Write( root.data + " ");
    inorderTraversal(root.right);
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = {7, 4, 12, 3,
                 6, 8, 1, 5, 10};
    int n = arr.Length;
     
    Node root = constructBst(arr, n);
     
    Console.Write("Inorder Traversal: ");
    inorderTraversal(root);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
      // JavaScript implementation to construct a BST
      // from its level order traversal
      // node of a BST
      class Node {
        constructor() {
          this.data = 0;
          this.left = null;
          this.right = null;
        }
      }
 
      // function to get a new node
      function getNode(data) {
        // Allocate memory
        var newNode = new Node();
 
        // put in the data
        newNode.data = data;
        newNode.left = newNode.right = null;
        return newNode;
      }
 
      // function to construct a BST from
      // its level order traversal
      function LevelOrder(root, data) {
        if (root == null) {
          root = getNode(data);
          return root;
        }
 
        if (data <= root.data)
        root.left = LevelOrder(root.left, data);
        else
        root.right = LevelOrder(root.right, data);
        return root;
      }
 
      function constructBst(arr, n) {
        if (n == 0) return null;
        var root = null;
 
        for (var i = 0; i < n; i++)
        root = LevelOrder(root, arr[i]);
 
        return root;
      }
 
      // function to print the inorder traversal
      function inorderTraversal(root) {
        if (root == null) return;
 
        inorderTraversal(root.left);
        document.write(root.data + " ");
        inorderTraversal(root.right);
      }
 
      // Driver code
      var arr = [7, 4, 12, 3, 6, 8, 1, 5, 10];
      var n = arr.length;
 
      var root = constructBst(arr, n);
 
      document.write("Inorder Traversal: ");
      inorderTraversal(root);
       
</script>

Output:  

Inorder Traversal: 1 3 4 5 6 7 8 10 12

Time Complexity : O(n2
This is because the above program is like we are inserting n nodes in a bst which takes O(n2) time in the worst case.
This article is contributed by Nishant Balayan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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