Convert BST to Min Heap
Given a binary search tree which is also a complete binary tree. The problem is to convert the given BST into a Min Heap with the condition that all the values in the left subtree of a node should be less than all the values in the right subtree of the node. This condition is applied to all the nodes, in the resultant converted Min Heap.
Examples:
Input: 4
/ \
2 6
/ \ / \
1 3 5 7
Output: 1
/ \
2 5
/ \ / \
3 4 6 7
Explanation: The given BST has been transformed into a Min Heap. All the nodes in the Min Heap satisfies the given condition, that is, values in the left subtree of a node should be less than the values in the right subtree of the node.
Approach: To solve the problem using this approach follow the below idea:
Store the inorder traversal of the BST in array and then do preorder traversal of the BST and while doing preorder traversal copy the values of inorder traversal into the current node, as copying the sorted elements while doing preorder traversal will make sure that a Min-Heap is constructed with the condition that all the values in the left subtree of a node are less than all the values in the right subtree of the node.
Follow the given steps to solve the problem:
- Create an array arr[] of size N, where N is the number of nodes in the given BST.
- Perform the inorder traversal of the BST and copy the node values in the arr[] in sorted order.
- Now perform the preorder traversal of the tree.
- While traversing the root during the preorder traversal, one by one copy the values from the array arr[] to the nodes of the BST.
Below is the implementation of the above approach:
C++
// C++ implementation to convert the given // BST to Min Heap #include <bits/stdc++.h> using namespace std; // Structure of a node of BST struct Node { int data; Node *left, *right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct Node* getNode( int data) { struct Node* newNode = new Node; newNode->data = data; newNode->left = newNode->right = NULL; return newNode; } // function prototype for preorder traversal // of the given tree void preorderTraversal(Node*); // function for the inorder traversal of the tree // so as to store the node values in 'arr' in // sorted order void inorderTraversal(Node* root, vector< int >& arr) { if (root == NULL) return ; // first recur on left subtree inorderTraversal(root->left, arr); // then copy the data of the node arr.push_back(root->data); // now recur for right subtree inorderTraversal(root->right, arr); } // function to convert the given BST to MIN HEAP // performs preorder traversal of the tree void BSTToMinHeap(Node* root, vector< int > arr, int * i) { if (root == NULL) return ; // first copy data at index 'i' of 'arr' to // the node root->data = arr[++*i]; // then recur on left subtree BSTToMinHeap(root->left, arr, i); // now recur on right subtree BSTToMinHeap(root->right, arr, i); } // utility function to convert the given BST to // MIN HEAP void convertToMinHeapUtil(Node* root) { // vector to store the data of all the // nodes of the BST vector< int > arr; int i = -1; // inorder traversal to populate 'arr' inorderTraversal(root, arr); // BST to MIN HEAP conversion BSTToMinHeap(root, arr, &i); } // function for the preorder traversal of the tree void preorderTraversal(Node* root) { if (!root) return ; // first print the root's data cout << root->data << " " ; // then recur on left subtree preorderTraversal(root->left); // now recur on right subtree preorderTraversal(root->right); } // Driver program to test above int main() { // BST formation struct Node* root = getNode(4); root->left = getNode(2); root->right = getNode(6); root->left->left = getNode(1); root->left->right = getNode(3); root->right->left = getNode(5); root->right->right = getNode(7); // Function call convertToMinHeapUtil(root); cout << "Preorder Traversal:" << endl; preorderTraversal(root); return 0; } |
Java
// Java implementation to convert the given // BST to Min Heap import java.util.ArrayList; class Gfg { static class Node { int data; Node left, right; // Constructor Node() { this .data = 0 ; this .left = this .right = null ; } Node( int data) { this .data = data; this .left = this .right = null ; } } private static void preOrder(Node root) { if (root == null ) return ; System.out.print(root.data + " " ); preOrder(root.left); preOrder(root.right); } private static void bstToArray(Node root, ArrayList<Integer> arr) { // ArrayLIst stores elements in inorder fashion if (root == null ) return ; bstToArray(root.left, arr); arr.add(root.data); bstToArray(root.right, arr); } static int index; private static void arrToMinHeap(Node root, ArrayList<Integer> arr) { if (root == null ) return ; root.data = arr.get(index++); arrToMinHeap(root.left, arr); arrToMinHeap(root.right, arr); } static void convertToMinHeap(Node root) { // initialize static index to zero index = 0 ; ArrayList<Integer> arr = new ArrayList<Integer>(); bstToArray(root, arr); arrToMinHeap(root, arr); } // Driver's code public static void main(String[] args) { // BST formation Node root = new Node( 4 ); root.left = new Node( 2 ); root.right = new Node( 6 ); root.left.left = new Node( 1 ); root.left.right = new Node( 3 ); root.right.left = new Node( 5 ); root.right.right = new Node( 7 ); System.out.print( "Preorder Traversal Before Conversion :" + "\n" ); preOrder(root); // Function call convertToMinHeap(root); System.out.print( "\nPreorder Traversal After Conversion :" + "\n" ); preOrder(root); } } // Contributed by : @mahi_07 /* Tip : If interviewer ask not to use global index variable you can use LinkedList Instead of ArrayList and use LinkedList's removeFirst() method So instead of this root.data = arr.get(index++); you can write root.data = list.removeFirst(); Do not forget to initialize list in converttoMinHeap function */ |
Python3
# Python3 implementation to convert the # given BST to Min Heap # structure of a node of BST class Node: # Constructor to create a new node def __init__( self , data): self .data = data self .left = None self .right = None # function for the inorder traversal # of the tree so as to store the node # values in 'arr' in sorted order def inorderTraversal(root, arr): if root = = None : return # first recur on left subtree inorderTraversal(root.left, arr) # then copy the data of the node arr.append(root.data) # now recur for right subtree inorderTraversal(root.right, arr) # function to convert the given # BST to MIN HEAP performs preorder # traversal of the tree def BSTToMinHeap(root, arr, i): if root = = None : return # first copy data at index 'i' of # 'arr' to the node i[ 0 ] + = 1 root.data = arr[i[ 0 ]] # then recur on left subtree BSTToMinHeap(root.left, arr, i) # now recur on right subtree BSTToMinHeap(root.right, arr, i) # utility function to convert the # given BST to MIN HEAP def convertToMinHeapUtil(root): # vector to store the data of # all the nodes of the BST arr = [] i = [ - 1 ] # inorder traversal to populate 'arr' inorderTraversal(root, arr) # BST to MIN HEAP conversion BSTToMinHeap(root, arr, i) # function for the preorder traversal # of the tree def preorderTraversal(root): if root = = None : return # first print the root's data print (root.data, end = " " ) # then recur on left subtree preorderTraversal(root.left) # now recur on right subtree preorderTraversal(root.right) # Driver's Code if __name__ = = '__main__' : # BST formation root = Node( 4 ) root.left = Node( 2 ) root.right = Node( 6 ) root.left.left = Node( 1 ) root.left.right = Node( 3 ) root.right.left = Node( 5 ) root.right.right = Node( 7 ) # Function call convertToMinHeapUtil(root) print ( "Preorder Traversal:" ) preorderTraversal(root) # This code is contributed # by PranchalK |
C#
// C# implementation to convert the given // BST to Min Heap using System; using System.Collections.Generic; public class GFG { // structure of a node of BST public class Node { public int data; public Node left, right; }; /* Helper function that allocates a new node with the given data and null left and right pointers. */ static Node getNode( int data) { Node newNode = new Node(); newNode.data = data; newNode.left = newNode.right = null ; return newNode; } // function prototype for preorder traversal // of the given tree // function for the inorder traversal of the tree // so as to store the node values in 'arr' in // sorted order static void inorderTraversal(Node root) { if (root == null ) return ; // first recur on left subtree inorderTraversal(root.left); // then copy the data of the node arr.Add(root.data); // now recur for right subtree inorderTraversal(root.right); } // function to convert the given BST to MIN HEAP // performs preorder traversal of the tree static void BSTToMinHeap(Node root) { if (root == null ) return ; // first copy data at index 'i' of 'arr' to // the node root.data = arr[++i]; // then recur on left subtree BSTToMinHeap(root.left); // now recur on right subtree BSTToMinHeap(root.right); } static List< int > arr = new List< int >(); static int i; // utility function to convert the given BST to // MIN HEAP static void convertToMinHeapUtil(Node root) { // vector to store the data of all the // nodes of the BST i = -1; // inorder traversal to populate 'arr' inorderTraversal(root); // BST to MIN HEAP conversion BSTToMinHeap(root); } // function for the preorder traversal of the tree static void preorderTraversal(Node root) { if (root == null ) return ; // first print the root's data Console.Write(root.data + " " ); // then recur on left subtree preorderTraversal(root.left); // now recur on right subtree preorderTraversal(root.right); } // Driver program to test above public static void Main(String[] args) { // BST formation Node root = getNode(4); root.left = getNode(2); root.right = getNode(6); root.left.left = getNode(1); root.left.right = getNode(3); root.right.left = getNode(5); root.right.right = getNode(7); convertToMinHeapUtil(root); Console.Write( "Preorder Traversal:" + "\n" ); preorderTraversal(root); } } // This code contributed by Rajput-Ji |
Javascript
// JavaScript implementation to convert the given // BST to Min Heap // structure of a node of BST class Node { constructor() { this .data = 0; this .left = null ; this .right = null ; } } /* Helper function that allocates a new node with the given data and null left and right pointers. */ function getNode(data) { var newNode = new Node(); newNode.data = data; newNode.left = newNode.right = null ; return newNode; } // function prototype for preorder traversal // of the given tree // function for the inorder traversal of the tree // so as to store the node values in 'arr' in // sorted order function inorderTraversal(root) { if (root == null ) return ; // first recur on left subtree inorderTraversal(root.left); // then copy the data of the node arr.push(root.data); // now recur for right subtree inorderTraversal(root.right); } // function to convert the given BST to MIN HEAP // performs preorder traversal of the tree function BSTToMinHeap(root) { if (root == null ) return ; // first copy data at index 'i' of 'arr' to // the node root.data = arr[++i]; // then recur on left subtree BSTToMinHeap(root.left); // now recur on right subtree BSTToMinHeap(root.right); } var arr = []; var i; // utility function to convert the given BST to // MIN HEAP function convertToMinHeapUtil(root) { // vector to store the data of all the // nodes of the BST i = -1; // inorder traversal to populate 'arr' inorderTraversal(root); // BST to MIN HEAP conversion BSTToMinHeap(root); } // function for the preorder traversal of the tree function preorderTraversal(root) { if (root == null ) { return ; } // first print the root's data document.write(root.data + " " ); // then recur on left subtree preorderTraversal(root.left); // now recur on right subtree preorderTraversal(root.right); } // Driver program to test above // BST formation var root = getNode(4); root.left = getNode(2); root.right = getNode(6); root.left.left = getNode(1); root.left.right = getNode(3); root.right.left = getNode(5); root.right.right = getNode(7); convertToMinHeapUtil(root); document.write( "Preorder Traversal: " ); preorderTraversal(root); |
Preorder Traversal: 1 2 3 4 5 6 7
Time Complexity: O(N)
Auxiliary Space: O(N)
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