Remove BST Keys in a given Range

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

There are two possible cases for every node.
1) Node’s key is inside the given range.
2) 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    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 node BSTnode* leftMost(BSTnode* root) {     if (!root)         return NULL;     while (root->left)         root = root->left;     return root; }    // A Utility function to delete the give node BSTnode* 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 manner BSTnode* 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 conversion void inorder(BSTnode* root) {     if (root) {         inorder(root->left);         cout << root->data << ' ';         inorder(root->right);     } }    // Driver Program to test above functions int 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 approach import 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 give 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

C#

 // C# implementation of the approach using 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 give 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

Output:

Inorder Before deletion: 20 30 40 50 60 70 80
Inorder After deletion: 20 30 40 80

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