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
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> |
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.
- Check whether arr[i] can be a left child or right child of the Node by checking the condition of BST.
- 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) { if (temp->left != NULL){ //if temp already has a left child //temp can have no more children q.pop(); i--; continue ; } // 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) |
Java
// Java code for the above approach import java.io.*; import java.util.*; // Node of a BST class Node { int data; Node left, right; public Node( int data) { this .data = data; this .left = null ; this .right = null ; } } public class GFG { static class NodeRange { Node node; int min, max; public NodeRange(Node node, int min, int max) { this .node = node; this .min = min; this .max = max; } } public static Node constructBst( int [] arr) { if (arr.length == 0 ) return null ; // Create root node and store a copy of it in head Node root = new Node(arr[ 0 ]), head = root; // Create queue to store the tree nodes Queue<NodeRange> queue = new LinkedList<>(); queue.add( new NodeRange(root, Integer.MIN_VALUE, Integer.MAX_VALUE)); for ( int i = 1 ; i < arr.length; i++) { NodeRange nr = queue.peek(); // Check if arr[i] can be a child of the temp // node if (arr[i] > nr.min && arr[i] < nr.max) { // Check if arr[i] can be left child if (arr[i] < nr.node.data) { // Set the left child and range nr.node.left = new Node(arr[i]); queue.add( new NodeRange(nr.node.left, nr.min, nr.node.data)); } // Check if arr[i] can be right child else { // Pop the temp node from queue, set the // right child and new range queue.remove(); nr.node.right = new Node(arr[i]); queue.add( new NodeRange(nr.node.right, nr.node.data, nr.max)); } } else { queue.remove(); i--; } } return head; } public static void inorderTraversal(Node root) { if (root == null ) return ; inorderTraversal(root.left); System.out.print(root.data + " " ); inorderTraversal(root.right); } public static void main(String[] args) { int [] arr = { 7 , 4 , 12 , 3 , 6 , 8 , 1 , 5 , 10 }; Node root = constructBst(arr); System.out.print( "Inorder Traversal: " ); inorderTraversal(root); } } // This code is contributed by sankar. |
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 |
C#
using System; using System.Collections.Generic; // Node of a BST class Node { public int data; public Node left; public Node right; public Node( int data) { this .data = data; left = null ; right = null ; } } class BST { public Node constructBst( int [] arr, int n) { var queue = new Queue<Tuple<Node, Tuple< int , int > > >(); if (n == 0) return null ; // Create root node and store a copy of it in head var root = new Node(arr[0]); var head = root; // Push the root node and the initial range queue.Enqueue( Tuple.Create(root, Tuple.Create( int .MinValue, int .MaxValue))); int 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 var temp = queue.Dequeue(); var tempRange = temp.Item2; // Check if arr[i] can be left child and within // range of it's parent data if (arr[i] < temp.Item1.data && tempRange.Item1 < arr[i] && arr[i] < tempRange.Item2) { // Set the left child and new range for the // child temp.Item1.left = new Node(arr[i]); queue.Enqueue(Tuple.Create( temp.Item1.left, Tuple.Create(tempRange.Item1, temp.Item1.data))); i++; } // Check if arr[i] can be right child and within // range of it's parent data if (arr[i] > temp.Item1.data && tempRange.Item1 < arr[i] && arr[i] < tempRange.Item2) { // Set the right child and new range for the // child temp.Item1.right = new Node(arr[i]); queue.Enqueue(Tuple.Create( temp.Item1.right, Tuple.Create(temp.Item1.data, tempRange.Item2))); i++; } } return head; } public void inorderTraversal(Node root) { if (root == null ) return ; inorderTraversal(root.left); Console.Write(root.data + " " ); inorderTraversal(root.right); } // Driver program public static void Main( string [] args) { var arr = new int [] { 7, 4, 12, 3, 6, 8, 1, 5, 10 }; int n = arr.Length; var bst = new BST(); var root = bst.constructBst(arr, n); Console.WriteLine( "Inorder Traversal:" ); bst.inorderTraversal(root); } } |
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. |
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
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