Construct BST from its given level order traversal

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.

This article is contributed by Nishant Balayan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

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

C++

Java

Python3

C#

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