Remove BST keys outside the given range
Given a Binary Search Tree (BST) and a range [min, max], remove all keys which are outside the given range. The modified tree should also be BST. For example, consider the following BST and range [-10, 13].
The given tree should be changed to the following. Note that all keys outside the range [-10, 13] are removed and the modified tree is BST.
There are two possible cases for every node.
1) Node’s key is outside the given range. This case has two sub-cases.
…….a) Node’s key is smaller than the min value.
…….b) Node’s key is greater than the max value.
2) Node’s key is in range.
We don’t need to do anything for case 2. In case 1, we need to remove the node and change the root of the subtree rooted with this node.
The idea is to fix the tree in a Post-order fashion. When we visit a node, we make sure that its left and right sub-trees are already fixed. In case 1.a), we simply remove the root and return the right sub-tree as a new root. In case 1.b), we remove the root and return the left sub-tree as a new root.
Following is the implementation of the above approach.
C++
// A C++ program to remove BST keys outside the given range#include<bits/stdc++.h>using namespace std;// A BST node has key, and left and right pointersstruct node{ int key; struct node *left; struct node *right;};// Removes all nodes having value outside the given range and returns the root// of modified treenode* removeOutsideRange(node *root, int min, int max){ // Base Case if (root == NULL) return NULL; // First fix the left and right subtrees of root root->left = removeOutsideRange(root->left, min, max); root->right = removeOutsideRange(root->right, min, max); // Now fix the root. There are 2 possible cases for root // 1.a) Root's key is smaller than min value (root is not in range) if (root->key < min) { node *rChild = root->right; delete root; return rChild; } // 1.b) Root's key is greater than max value (root is not in range) if (root->key > max) { node *lChild = root->left; delete root; return lChild; } // 2. Root is in range return root;}// A utility function to create a new BST node with key as given numnode* newNode(int num){ node* temp = new node; temp->key = num; temp->left = temp->right = NULL; return temp;}// A utility function to insert a given key to BSTnode* insert(node* root, int key){ if (root == NULL) return newNode(key); if (root->key > key) root->left = insert(root->left, key); else root->right = insert(root->right, key); return root;}// Utility function to traverse the binary tree after conversionvoid inorderTraversal(node* root){ if (root) { inorderTraversal( root->left ); cout << root->key << " "; inorderTraversal( root->right ); }}// Driver program to test above functionsint main(){ node* root = NULL; root = insert(root, 6); root = insert(root, -13); root = insert(root, 14); root = insert(root, -8); root = insert(root, 15); root = insert(root, 13); root = insert(root, 7); cout << "Inorder traversal of the given tree is: "; inorderTraversal(root); root = removeOutsideRange(root, -10, 13); cout << "\nInorder traversal of the modified tree is: "; inorderTraversal(root); return 0;} |
Java
// A Java program to remove BST// keys outside the given rangeimport java.math.BigDecimal;import java.util.ArrayList;import java.util.Arrays;import java.util.List;import java.util.Scanner;class Node{ int key; Node left; Node right;}class GFG{ // Removes all nodes having value // outside the given range and // returns the root of modified tree private static Node removeOutsideRange(Node root, int min, int max) { // BASE CASE if(root == null) { return null; } // FIRST FIX THE LEFT AND // RIGHT SUBTREE OF ROOT root.left = removeOutsideRange(root.left, min, max); root.right = removeOutsideRange(root.right, min, max); // NOW FIX THE ROOT. THERE ARE // TWO POSSIBLE CASES FOR THE ROOT // 1. a) Root's key is smaller than // min value(root is not in range) if(root.key < min) { Node rchild = root.right; root = null; return rchild; } // 1. b) Root's key is greater than // max value (Root is not in range) if(root.key > max) { Node lchild = root.left; root = null; return lchild; } // 2. Root in range return root; } public static Node newNode(int num) { Node temp = new Node(); temp.key = num; temp.left = null; temp.right = null; return temp; } public static Node insert(Node root, int key) { if(root == null) { return newNode(key); } if(root.key > key) { root.left = insert(root.left, key); } else { root.right = insert(root.right, key); } return root; } private static void inorderTraversal(Node root) { if(root != null) { inorderTraversal(root.left); System.out.print(root.key + " "); inorderTraversal(root.right); } } // Driver code public static void main(String[] args) { Node root = null; root = insert(root, 6); root = insert(root, -13); root = insert(root, 14); root = insert(root, -8); root = insert(root, 15); root = insert(root, 13); root = insert(root, 7); System.out.print("Inorder Traversal of " + "the given tree is: "); inorderTraversal(root); root = removeOutsideRange(root, -10, 13); System.out.print("\nInorder traversal of " + "the modified tree: "); inorderTraversal(root); }}// This code is contributed // by Divya |
Python3
# Python3 program to remove BST keys # outside the given range # A BST node has key, and left and right # pointers. A utility function to create # a new BST node with key as given num class newNode: # Constructor to create a new node def __init__(self, data): self.key = data self.left = None self.right = None # Removes all nodes having value outside # the given range and returns the root # of modified tree def removeOutsideRange(root, Min, Max): # Base Case if root == None: return None # First fix the left and right # subtrees of root root.left = removeOutsideRange(root.left, Min, Max) root.right = removeOutsideRange(root.right, Min, Max) # Now fix the root. There are 2 # possible cases for root # 1.a) Root's key is smaller than # min value (root is not in range) if root.key < Min: rChild = root.right return rChild # 1.b) Root's key is greater than max # value (root is not in range) if root.key > Max: lChild = root.left return lChild # 2. Root is in range return root# A utility function to insert a given# key to BST def insert(root, key): if root == None: return newNode(key) if root.key > key: root.left = insert(root.left, key) else: root.right = insert(root.right, key) return root# Utility function to traverse the binary# tree after conversion def inorderTraversal(root): if root: inorderTraversal( root.left) print(root.key, end = " ") inorderTraversal( root.right)# Driver Code if __name__ == '__main__': root = None root = insert(root, 6) root = insert(root, -13) root = insert(root, 14) root = insert(root, -8) root = insert(root, 15) root = insert(root, 13) root = insert(root, 7) print("Inorder traversal of the given tree is:", end = " ") inorderTraversal(root) root = removeOutsideRange(root, -10, 13) print() print("Inorder traversal of the modified tree is:", end = " ") inorderTraversal(root) # This code is contributed by PranchalK |
C#
// A C# program to remove BST// keys outside the given rangeusing System;public class Node { public int key; public Node left; public Node right; } public class GFG { // Removes all nodes having value // outside the given range and // returns the root of modified tree private static Node removeOutsideRange(Node root, int min, int max) { // BASE CASE if(root == null) { return null; } // FIRST FIX THE LEFT AND // RIGHT SUBTREE OF ROOT root.left = removeOutsideRange(root.left, min, max); root.right = removeOutsideRange(root.right, min, max); // NOW FIX THE ROOT. THERE ARE // TWO POSSIBLE CASES FOR THE ROOT // 1. a) Root's key is smaller than // min value(root is not in range) if(root.key < min) { Node rchild = root.right; root = null; return rchild; } // 1. b) Root's key is greater than // max value (Root is not in range) if(root.key > max) { Node lchild = root.left; root = null; return lchild; } // 2. Root in range return root; } public static Node newNode(int num) { Node temp = new Node(); temp.key = num; temp.left = null; temp.right = null; return temp; } public static Node insert(Node root, int key) { if(root == null) { return newNode(key); } if(root.key > key) { root.left = insert(root.left, key); } else { root.right = insert(root.right, key); } return root; } private static void inorderTraversal(Node root) { if(root != null) { inorderTraversal(root.left); Console.Write(root.key + " "); inorderTraversal(root.right); } } // Driver code public static void Main(String[] args) { Node root = null; root = insert(root, 6); root = insert(root, -13); root = insert(root, 14); root = insert(root, -8); root = insert(root, 15); root = insert(root, 13); root = insert(root, 7); Console.Write("Inorder Traversal of " + "the given tree is: "); inorderTraversal(root); root = removeOutsideRange(root, -10, 13); Console.Write("\nInorder traversal of " + "the modified tree: "); inorderTraversal(root); } } // This code has been contributed // by PrinciRaj1992 |
Output:
Inorder traversal of the given tree is: -13 -8 6 7 13 14 15 Inorder traversal of the modified tree is: -8 6 7 13
Time Complexity: O(n) where n is the number of nodes in a given BST.
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