Construct BST from its given level order traversal

Difficulty Level : Hard
Last Updated : 10 Jun, 2020

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

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.

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