Print the Forests of a Binary Tree after removal of given Nodes Last Updated : 15 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a Binary Tree and an array arr[] consisting of values of nodes to be deleted, the task is to print Inorder Traversal of the forests after deleting the nodes. Examples: Input: arr[] = {10, 5} 10 / \ 20 30 / \ \ 4 5 7 Output: 4 20 30 7Input: arr[] = {5} 1 / \ 5 6 / \ 10 12 Output: 10 1 6 12 Approach: Follow the below steps to solve the problem: Perform the Postorder Traversal of the Binary Tree.For each node, check if it contains the value to be deleted.If found to be true, store its child as the root of the forest. Print the forest by Inorder Traversal. Below is the implementation of the above approach: C++ // C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Stores the nodes to be deleted unordered_map<int, bool> mp; // Structure of a Tree node struct Node { int key; struct Node *left, *right; }; // Function to create a new node Node* newNode(int key) { Node* temp = new Node; temp->key = key; temp->left = temp->right = NULL; return (temp); } // Function to check whether the node // needs to be deleted or not bool deleteNode(int nodeVal) { return mp.find(nodeVal) != mp.end(); } // Function to perform tree pruning // by performing post order traversal Node* treePruning(Node* root, vector<Node*>& result) { if (root == NULL) return NULL; root->left = treePruning(root->left, result); root->right = treePruning(root->right, result); // If the node needs to be deleted if (deleteNode(root->key)) { // Store the its subtree if (root->left) { result.push_back(root->left); } if (root->right) { result.push_back(root->right); } return NULL; } return root; } // Perform Inorder Traversal void printInorderTree(Node* root) { if (root == NULL) return; printInorderTree(root->left); cout << root->key << " "; printInorderTree(root->right); } // Function to print the forests void printForests(Node* root, int arr[], int n) { for (int i = 0; i < n; i++) { mp[arr[i]] = true; } // Stores the remaining nodes vector<Node*> result; if (treePruning(root, result)) result.push_back(root); // Print the inorder traversal of Forests for (int i = 0; i < result.size(); i++) { printInorderTree(result[i]); cout << endl; } } // Driver Code int main() { Node* root = newNode(1); root->left = newNode(12); root->right = newNode(13); root->right->left = newNode(14); root->right->right = newNode(15); root->right->left->left = newNode(21); root->right->left->right = newNode(22); root->right->right->left = newNode(23); root->right->right->right = newNode(24); int arr[] = { 14, 23, 1 }; int n = sizeof(arr) / sizeof(arr[0]); printForests(root, arr, n); } Java // Java Program to implement // the above approach import java.util.*; class GFG{ // Stores the nodes to be deleted static HashMap<Integer, Boolean> mp = new HashMap<>(); // Structure of a Tree node static class Node { int key; Node left, right; }; // Function to create a new node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.left = temp.right = null; return (temp); } // Function to check whether the node // needs to be deleted or not static boolean deleteNode(int nodeVal) { return mp.containsKey(nodeVal); } // Function to perform tree pruning // by performing post order traversal static Node treePruning(Node root, Vector<Node> result) { if (root == null) return null; root.left = treePruning(root.left, result); root.right = treePruning(root.right, result); // If the node needs to be deleted if (deleteNode(root.key)) { // Store the its subtree if (root.left != null) { result.add(root.left); } if (root.right != null) { result.add(root.right); } return null; } return root; } // Perform Inorder Traversal static void printInorderTree(Node root) { if (root == null) return; printInorderTree(root.left); System.out.print(root.key + " "); printInorderTree(root.right); } // Function to print the forests static void printForests(Node root, int arr[], int n) { for (int i = 0; i < n; i++) { mp.put(arr[i], true); } // Stores the remaining nodes Vector<Node> result = new Vector<>(); if (treePruning(root, result) != null) result.add(root); // Print the inorder traversal of Forests for (int i = 0; i < result.size(); i++) { printInorderTree(result.get(i)); System.out.println(); } } // Driver Code public static void main(String[] args) { Node root = newNode(1); root.left = newNode(12); root.right = newNode(13); root.right.left = newNode(14); root.right.right = newNode(15); root.right.left.left = newNode(21); root.right.left.right = newNode(22); root.right.right.left = newNode(23); root.right.right.right = newNode(24); int arr[] = { 14, 23, 1 }; int n = arr.length; printForests(root, arr, n); } } // This code is contributed by Rajput-Ji Python3 # Python3 Program to implement # the above approach # Stores the nodes to be deleted mp = dict() # Structure of a Tree node class Node: def __init__(self, key): self.key = key self.left = None self.right = None # Function to create a new node def newNode(key): temp = Node(key) return temp # Function to check whether the node # needs to be deleted or not def deleteNode(nodeVal): if nodeVal in mp: return True else: return False # Function to perform tree pruning # by performing post order traversal def treePruning( root, result): if (root == None): return None; root.left = treePruning(root.left, result); root.right = treePruning(root.right, result); # If the node needs to be deleted if (deleteNode(root.key)): # Store the its subtree if (root.left): result.append(root.left); if (root.right): result.append(root.right); return None; return root; # Perform Inorder Traversal def printInorderTree(root): if (root == None): return; printInorderTree(root.left); print(root.key, end=' ') printInorderTree(root.right); # Function to print the forests def printForests(root, arr, n): for i in range(n): mp[arr[i]] = True; # Stores the remaining nodes result = [] if (treePruning(root, result)): result.append(root); # Print the inorder traversal of Forests for i in range(len(result)): printInorderTree(result[i]); print() # Driver Code if __name__=='__main__': root = newNode(1); root.left = newNode(12); root.right = newNode(13); root.right.left = newNode(14); root.right.right = newNode(15); root.right.left.left = newNode(21); root.right.left.right = newNode(22); root.right.right.left = newNode(23); root.right.right.right = newNode(24); arr = [ 14, 23, 1 ] n = len(arr) printForests(root, arr, n); # This code is contributed by rutvik_56 C# // C# program to implement // the above approach using System; using System.Collections.Generic; class GFG{ // Stores the nodes to be deleted static Dictionary<int, Boolean> mp = new Dictionary<int, Boolean>(); // Structure of a Tree node class Node { public int key; public Node left, right; }; // Function to create a new node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.left = temp.right = null; return (temp); } // Function to check whether the node // needs to be deleted or not static bool deleteNode(int nodeVal) { return mp.ContainsKey(nodeVal); } // Function to perform tree pruning // by performing post order traversal static Node treePruning(Node root, List<Node> result) { if (root == null) return null; root.left = treePruning(root.left, result); root.right = treePruning(root.right, result); // If the node needs to be deleted if (deleteNode(root.key)) { // Store the its subtree if (root.left != null) { result.Add(root.left); } if (root.right != null) { result.Add(root.right); } return null; } return root; } // Perform Inorder Traversal static void printInorderTree(Node root) { if (root == null) return; printInorderTree(root.left); Console.Write(root.key + " "); printInorderTree(root.right); } // Function to print the forests static void printForests(Node root, int []arr, int n) { for(int i = 0; i < n; i++) { mp.Add(arr[i], true); } // Stores the remaining nodes List<Node> result = new List<Node>(); if (treePruning(root, result) != null) result.Add(root); // Print the inorder traversal of Forests for(int i = 0; i < result.Count; i++) { printInorderTree(result[i]); Console.WriteLine(); } } // Driver Code public static void Main(String[] args) { Node root = newNode(1); root.left = newNode(12); root.right = newNode(13); root.right.left = newNode(14); root.right.right = newNode(15); root.right.left.left = newNode(21); root.right.left.right = newNode(22); root.right.right.left = newNode(23); root.right.right.right = newNode(24); int []arr = { 14, 23, 1 }; int n = arr.Length; printForests(root, arr, n); } } // This code is contributed by Amit Katiyar JavaScript <script> // Javascript Program to implement the above approach // Stores the nodes to be deleted let mp = new Map(); // Structure of a Tree node class Node { constructor(key) { this.left = this.right = null; this.key = key; } } // Function to create a new node function newNode(key) { let temp = new Node(key); return (temp); } // Function to check whether the node // needs to be deleted or not function deleteNode(nodeVal) { return mp.has(nodeVal); } // Function to perform tree pruning // by performing post order traversal function treePruning(root, result) { if (root == null) return null; root.left = treePruning(root.left, result); root.right = treePruning(root.right, result); // If the node needs to be deleted if (deleteNode(root.key)) { // Store the its subtree if (root.left != null) { result.push(root.left); } if (root.right != null) { result.push(root.right); } return null; } return root; } // Perform Inorder Traversal function printInorderTree(root) { if (root == null) return; printInorderTree(root.left); document.write(root.key + " "); printInorderTree(root.right); } // Function to print the forests function printForests(root, arr, n) { for (let i = 0; i < n; i++) { mp.set(arr[i], true); } // Stores the remaining nodes let result = []; if (treePruning(root, result) != null) result.push(root); // Print the inorder traversal of Forests for (let i = 0; i < result.length; i++) { printInorderTree(result[i]); document.write("</br>"); } } let root = newNode(1); root.left = newNode(12); root.right = newNode(13); root.right.left = newNode(14); root.right.right = newNode(15); root.right.left.left = newNode(21); root.right.left.right = newNode(22); root.right.right.left = newNode(23); root.right.right.right = newNode(24); let arr = [ 14, 23, 1 ]; let n = arr.length; printForests(root, arr, n); </script> Output: 21 22 12 13 15 24 Time Complexity: O(N)Auxiliary Space: O(N) Comment More infoAdvertise with us Next Article Analysis of Algorithms A arpit7714 Follow Improve Article Tags : DSA Inorder Traversal PostOrder Traversal Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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