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
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 BSTstruct Node{ int data; Node *left, *right;};// function to get a new nodeNode* 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 traversalNode *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 traversalvoid inorderTraversal(Node* root){ if (!root) return; inorderTraversal(root->left); cout << root->data << " "; inorderTraversal(root->right); }// Driver program to test aboveint 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 traversalclass GFG{// node of a BSTstatic class Node{ int data; Node left, right;};// function to get a new nodestatic 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 traversalstatic 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 traversalstatic void inorderTraversal(Node root){ if (root == null) return; inorderTraversal(root.left); System.out.print( root.data + " "); inorderTraversal(root.right);}// Driver codepublic 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 traversalimport math# node of a BSTclass Node: def __init__(self,data): self.data = data self.left = None self.right = None# function to get a new nodedef 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 traversaldef 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 traversaldef inorderTraversal( root): if (root == None): return None inorderTraversal(root.left) print(root.data,end = " ") inorderTraversal(root.right)# Driver programif __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 traversalusing System;class GFG{// node of a BSTpublic class Node{ public int data; public Node left, right;};// function to get a new nodestatic 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 traversalstatic 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 traversalstatic void inorderTraversal(Node root){ if (root == null) return; inorderTraversal(root.left); Console.Write( root.data + " "); inorderTraversal(root.right);}// Driver codepublic 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.
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