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Construct BST from its given level order traversal

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  • Difficulty Level : Hard
  • Last Updated : 12 Jan, 2023
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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

Construct BST from its given level order traversal Using Recursion:

The idea is to use recursion as the first element will always be the root of the tree and second element will be the left child and the third element will be the right child (if fall in the range), and so on for all the remaining elements.

Follow the steps below to solve the problem:

  • First, pick the first element of the array and make it root. 
  • Pick the second element, if its value is smaller than the root node value make it left child, 
  • Else make it right child 
  • Now recursively call step (2) and step (3) to make a BST from its level Order Traversal.

Below is the implementation of the 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
 
import java.io.*;
 
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(N * H), Where N is the number of nodes in the tree and H is the height of the tree 
Auxiliary Space: O(N), N is the number of nodes in the Tree

Construct BST from its given level order traversal Using Queue:

The idea is similar to what we do while finding the level order traversal of a binary tree using the queue. In this case, we maintain a queue that contains a pair of the Node class and an integer pair storing the range for each of the tree nodes.

Follow the below steps to Implement the above idea:

  • Create an empty queue q<pair<Node*,pair<int,int>>> and push root and range from – infinite to  + infinite in q.
  • Run for loop through the entire array containing the level order traversal
  • Get the front of the queue and store its Node (in temp variable) and its range.
  • If arr[i] can be a child of temp by checking the value is within the range. 
    • Check whether arr[i] can be a left child or right child of the Node by checking the condition of BST.
      • If arr[i] can be a left child, we create a new Node and point it to the left child of temp. 
      • We update the range such that its new lower bound is the same as before and it’s new upper bound is the value of temp node.
      • If arr[i] can be the right child, we create a new Node and point it to the right child of temp. 
      • We update the range such that it’s new lower bound is the value of temp node and its new upper bound is the same as before. 
      • Pop the temp node from the queue once the right child is set. This is because the temp node cannot have any more children.
  • Else we pop out the node from the queue, decrement i and go ahead.
    • Initialize temp_node = q.front() and print temp_node->data.
    • Push temp_node’s children i.e. temp_node -> left then temp_node -> right to q
    • Pop front node from q.
  • Finally, return the head of the tree.

Below is the Implementation of the 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;
 
    Node(int x)
    {
        data = x;
        right = NULL;
        left = NULL;
    }
};
 
// Function to construct a BST from
// its level order traversal
Node* constructBst(int arr[], int n)
{
    // Create queue to store the tree nodes
    queue<pair<Node*, pair<int, int> > > q;
 
    // If array is empty we return NULL
    if (n == 0)
        return NULL;
 
    // Create root node and store a copy of it in head
    Node *root = new Node(arr[0]), *head = root;
 
    // Push the root node and the initial range
    q.push({ root, { INT_MIN, INT_MAX } });
 
    // Loop over the contents of arr to process all the
    // elements
    for (int i = 1; i < n; i++) {
 
        // Get the node and the range at the front of the
        // queue
        Node* temp = q.front().first;
        pair<int, int> range = q.front().second;
 
        // Check if arr[i] can be a child of the temp node
        if (arr[i] > range.first && arr[i] < range.second) {
 
            // Check if arr[i] can be left child
            if (arr[i] < temp->data) {
 
                // Set the left child and range
                temp->left = new Node(arr[i]);
                q.push({ temp->left,
                         { range.first, temp->data } });
            }
            // Check if arr[i] can be left child
            else {
 
                // Pop the temp node from queue, set the
                // right child and new range
                q.pop();
                temp->right = new Node(arr[i]);
                q.push({ temp->right,
                         { temp->data, range.second } });
            }
        }
        else {
 
            q.pop();
            i--;
        }
    }
    return head;
}
 
// 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;
}
 
// This code is contributed by Rohit Iyer (rohit_iyer)


Python3




# Python implementation to construct a BST
# from its level order traversal
 
# Importing essential libraries
from collections import deque
 
# Node of a BST
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
 
def constructBst(arr, n):
    queue = deque()
    if n == 0:
        return None
 
    # Create root node and store a copy of it in head
    root = Node(arr[0])
    head = root
 
    # Push the root node and the initial range
    queue.append((root, (-float("inf"), float("inf"))))
    i = 1
 
    # Loop over the contents of arr to process all the elements using
    # while loop we may have to process atmost 2 child's
    while i < n:
 
        # Get the node and the range at the front of the queue
        # and popout the leftmost element
        temp = queue[0][0]
        tempRange = queue[0][1]
        queue.popleft()
 
        # Check if arr[i] can be left child and within range of it's parent data
        if (arr[i] < temp.data) and tempRange[0] < arr[i] < tempRange[1]:
 
            # Set the left child and new range for the child
            temp.left = Node(arr[i])
            queue.append((temp.left, (tempRange[0], temp.data)))
            i += 1
 
        # Check if arr[i] can be right child and within range of it's parent data
        if arr[i] > temp.data and tempRange[0] < arr[i] < tempRange[1]:
 
            # Set the right child and new range for the child
            temp.right = Node(arr[i])
            queue.append((temp.right, (temp.data, tempRange[1])))
            i += 1
    return head
 
 
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: ")
    root = inorderTraversal(root)
 
 
# This code is contributed by Divyanshu Singh


Javascript




// Javascript code for the above approach
class Node {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
function constructBst(arr) {
    if (arr.length === 0) return null;
 
    // Create root node and store a copy of it in head
    let root = new Node(arr[0]), head = root;
 
    // Create queue to store the tree nodes
    let queue = [{ node: root, range: { min: Number.MIN_SAFE_INTEGER, max: Number.MAX_SAFE_INTEGER } }];
 
    for (let i = 1; i < arr.length; i++) {
        let { node, range } = queue[0];
 
        // Check if arr[i] can be a child of the temp node
        if (arr[i] > range.min && arr[i] < range.max) {
            // Check if arr[i] can be left child
            if (arr[i] < node.data) {
                // Set the left child and range
                node.left = new Node(arr[i]);
                queue.push({ node: node.left, range: { min: range.min, max: node.data } });
            }
            // Check if arr[i] can be left child
            else {
                // Pop the temp node from queue, set the right child and new range
                queue.shift();
                node.right = new Node(arr[i]);
                queue.push({ node: node.right, range: { min: node.data, max: range.max } });
            }
        }
        else {
            queue.shift();
            i--;
        }
    }
 
    return head;
}
 
function inorderTraversal(root) {
    if (!root) return;
    inorderTraversal(root.left);
    console.log(root.data+" ");
    inorderTraversal(root.right);
}
 
let arr = [7, 4, 12, 3, 6, 8, 1, 5, 10];
let root = constructBst(arr);
 
console.log("Inorder Traversal: ");
inorderTraversal(root);
 
// This code is contributed by lokeshpotta20.


Output

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

Time Complexity: O(N), Visiting every node once.
Auxiliary Space: O(N), Using queue to store the nodes

This article is contributed by Nishant Balayan and Rohit Iyer. 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. 


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