Given a Binary Search Tree (BST) and a range [min, max], remove all keys which are inside the given range. The modified tree should also be BST. For example, consider the following BST and range [50, 70].
50
/ \
30 70
/ \ / \
20 40 60 80
The given BST should be transformed to this:
80
/
30
/ \
20 40
There are two possible cases for every node.
- Node’s key is inside the given range.
- Node’s key is out of 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 sub-tree rooted with this node.
The idea is to fix the tree in Postorder fashion. When we visit a node, we make sure that its left and right sub-trees are already fixed. When we find a node inside the range we call normal BST delete function to delete that node.
Following is the implementation of the above approach.
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;class BSTnode {public: int data; BSTnode *left, *right; BSTnode(int data) { this->data = data; this->left = this->right = NULL; }};// A Utility function to find leftMost nodeBSTnode* leftMost(BSTnode* root){ if (!root) return NULL; while (root->left) root = root->left; return root;}// A Utility function to delete the given nodeBSTnode* deleteNode(BSTnode* root){ // node with only one chile or no child if (!root->left) { BSTnode* child = root->right; root = NULL; return child; } else if (!root->right) { BSTnode* child = root->left; root = NULL; return child; } // node with two children: get inorder successor // in the right subtree BSTnode* next = leftMost(root->right); // copy the inorder successor's content to this node root->data = next->data; // delete the inorder successor root->right = deleteNode(root->right); return root;}// function to find node in given range and delete// it in preorder mannerBSTnode* removeRange(BSTnode* node, int low, int high){ // Base case if (!node) return NULL; // First fix the left and right subtrees of node node->left = removeRange(node->left, low, high); node->right = removeRange(node->right, low, high); // Now fix the node. // if given node is in Range then delete it if (node->data >= low && node->data <= high) return deleteNode(node); // Root is out of range return node;}// Utility function to traverse the binary tree// after conversionvoid inorder(BSTnode* root){ if (root) { inorder(root->left); cout << root->data << ' '; inorder(root->right); }}// Driver Program to test above functionsint main(){ /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ BSTnode* root = new BSTnode(50); root->left = new BSTnode(30); root->right = new BSTnode(70); root->left->right = new BSTnode(40); root->right->right = new BSTnode(80); root->right->left = new BSTnode(60); root->left->left = new BSTnode(20); cout << "Inorder Before deletion: "; inorder(root); root = removeRange(root, 50, 70); cout << "\nInorder After deletion: "; inorder(root); cout << endl; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class Solution { static class BSTnode { int data; BSTnode left, right; BSTnode(int data) { this.data = data; this.left = this.right = null; } } // A Utility function to find leftMost node static BSTnode leftMost(BSTnode root) { if (root == null) return null; while (root.left != null) root = root.left; return root; } // A Utility function to delete the given node static BSTnode deleteNode(BSTnode root) { // node with only one chile or no child if (root.left == null) { BSTnode child = root.right; root = null; return child; } else if (root.right == null) { BSTnode child = root.left; root = null; return child; } // node with two children: get inorder successor // in the right subtree BSTnode next = leftMost(root.right); // copy the inorder successor's content to this node root.data = next.data; // delete the inorder successor root.right = deleteNode(root.right); return root; } // function to find node in given range and delete // it in preorder manner static BSTnode removeRange(BSTnode node, int low, int high) { // Base case if (node == null) return null; // First fix the left and right subtrees of node node.left = removeRange(node.left, low, high); node.right = removeRange(node.right, low, high); // Now fix the node. // if given node is in Range then delete it if (node.data >= low && node.data <= high) return deleteNode(node); // Root is out of range return node; } // Utility function to traverse the binary tree // after conversion static void inorder(BSTnode root) { if (root != null) { inorder(root.left); System.out.print(root.data + " "); inorder(root.right); } } // Driver Program to test above functions public static void main(String args[]) { /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ BSTnode root = new BSTnode(50); root.left = new BSTnode(30); root.right = new BSTnode(70); root.left.right = new BSTnode(40); root.right.right = new BSTnode(80); root.right.left = new BSTnode(60); root.left.left = new BSTnode(20); System.out.print("Inorder Before deletion: "); inorder(root); root = removeRange(root, 50, 70); System.out.print("\nInorder After deletion: "); inorder(root); }}// This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the above approachclass BSTnode: def __init__(self, data): self.data = data self.left = None self.right = None# A Utility function to find leftMost nodedef leftMost(root): if (root == None): return None while (root.left): root = root.left return root# A Utility function to delete the given nodedef deleteNode(root): # Node with only one chile or no child if (root.left == None): child = root.right root = None return child elif (root.right == None): child = root.left root = None return child # Node with two children: get # inorder successor in the # right subtree next = leftMost(root.right) # Copy the inorder successor's # content to this node root.data = next.data # Delete the inorder successor root.right = deleteNode(root.right) return root# Function to find node in given range # and delete it in preorder mannerdef removeRange(node, low, high): # Base case if (node == None): return None # First fix the left and right subtrees of node node.left = removeRange(node.left, low, high) node.right = removeRange(node.right, low, high) # Now fix the node. If given node # is in Range then delete it if (node.data >= low and node.data <= high): return deleteNode(node) # Root is out of range return node# Utility function to traverse the# binary tree after conversiondef inorder(root): if (root): inorder(root.left) print(root.data, end = " ") inorder(root.right)# Driver codeif __name__ == '__main__': ''' Let us create following BST 50 / \ 30 70 / / / \ 20 40 60 80 ''' root = BSTnode(50) root.left = BSTnode(30) root.right = BSTnode(70) root.left.right = BSTnode(40) root.right.right = BSTnode(80) root.right.left = BSTnode(60) root.left.left = BSTnode(20) print("Inorder Before deletion:", end = "") inorder(root) root = removeRange(root, 50, 70) print("\nInorder After deletion:", end = "") inorder(root)# This code is contributed by bgangwar59 |
C#
// C# implementation of the above approachusing System;class GFG { public class BSTnode { public int data; public BSTnode left, right; public BSTnode(int data) { this.data = data; this.left = this.right = null; } } // A Utility function to find leftMost node static BSTnode leftMost(BSTnode root) { if (root == null) return null; while (root.left != null) root = root.left; return root; } // A Utility function to delete the given node static BSTnode deleteNode(BSTnode root) { // node with only one chile or no child if (root.left == null) { BSTnode child = root.right; root = null; return child; } else if (root.right == null) { BSTnode child = root.left; root = null; return child; } // node with two children: get inorder successor // in the right subtree BSTnode next = leftMost(root.right); // copy the inorder successor's content to this node root.data = next.data; // delete the inorder successor root.right = deleteNode(root.right); return root; } // function to find node in given range and delete // it in preorder manner static BSTnode removeRange(BSTnode node, int low, int high) { // Base case if (node == null) return null; // First fix the left and right subtrees of node node.left = removeRange(node.left, low, high); node.right = removeRange(node.right, low, high); // Now fix the node. // if given node is in Range then delete it if (node.data >= low && node.data <= high) return deleteNode(node); // Root is out of range return node; } // Utility function to traverse the binary tree // after conversion static void inorder(BSTnode root) { if (root != null) { inorder(root.left); Console.Write(root.data + " "); inorder(root.right); } } // Driver code public static void Main(String[] args) { /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ BSTnode root = new BSTnode(50); root.left = new BSTnode(30); root.right = new BSTnode(70); root.left.right = new BSTnode(40); root.right.right = new BSTnode(80); root.right.left = new BSTnode(60); root.left.left = new BSTnode(20); Console.Write("Inorder Before deletion: "); inorder(root); root = removeRange(root, 50, 70); Console.Write("\nInorder After deletion: "); inorder(root); }}// This code contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the// above approachclass BSTnode{ constructor(data) { this.left = null; this.right = null; this.data = data; }}// A Utility function to find leftMost nodefunction leftMost(root){ if (root == null) return null; while (root.left != null) root = root.left; return root;}// A Utility function to delete the given nodefunction deleteNode(root){ // Node with only one chile or no child if (root.left == null) { let child = root.right; root = null; return child; } else if (root.right == null) { let child = root.left; root = null; return child; } // Node with two children: get inorder // successor in the right subtree let next = leftMost(root.right); // Copy the inorder successor's // content to this node root.data = next.data; // Delete the inorder successor root.right = deleteNode(root.right); return root;}// Function to find node in given range // and delete it in preorder mannerfunction removeRange(node, low, high){ // Base case if (node == null) return null; // First fix the left and right subtrees of node node.left = removeRange(node.left, low, high); node.right = removeRange(node.right, low, high); // Now fix the node. // if given node is in Range then delete it if (node.data >= low && node.data <= high) return deleteNode(node); // Root is out of range return node;}// Utility function to traverse the binary // tree after conversionfunction inorder(root){ if (root != null) { inorder(root.left); document.write(root.data + " "); inorder(root.right); }}// Driver code/* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */let root = new BSTnode(50);root.left = new BSTnode(30);root.right = new BSTnode(70);root.left.right = new BSTnode(40);root.right.right = new BSTnode(80);root.right.left = new BSTnode(60);root.left.left = new BSTnode(20);document.write("Inorder Before deletion: ");inorder(root);root = removeRange(root, 50, 70);document.write("</br>" + "Inorder After deletion: ");inorder(root);// This code is contributed by divyeshrabadiya07</script> |
Output
Inorder Before deletion: 20 30 40 50 60 70 80 Inorder After deletion: 20 30 40 80
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