Clone a Binary Tree with Random Pointers Last Updated : 29 Oct, 2024 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a Binary Tree where every node has data, a next pointer, a right pointer, and a random pointer. The random pointer points to any random node of the binary tree and can even point to NULL, the task is to clone the given binary tree.Example: Approach:The idea is to use hashmap to store mapping from given tree nodes to clone tree nodes in the hashtable. Then in second traversal assign each random pointer according to hash map.Follow the steps below to solve the problem:Traverse the original tree and create a new node for each original node, copying the data, left, and right pointers.Use a hashmap to store the mapping between original nodes and cloned nodes.Traverse the original tree again, using the hashmap to set the random pointers in the cloned nodes.Return the root of the newly cloned tree.Below is the implementation of above approach: C++ // A hashmap-based C++ program to clone a binary // tree with random pointers #include <iostream> #include <unordered_map> using namespace std; class Node { public: int data; Node *left; Node *right; Node *random; Node(int x) { data = x; left = right = random = nullptr; } }; // This function clones the tree by copying the // key (data) and left/right pointers. Node *copyLeftRightNode(Node *treeNode, unordered_map<Node *, Node *> &mp) { if (treeNode == nullptr) return nullptr; Node *cloneNode = new Node(treeNode->data); // Store the mapping between original node and cloned node mp[treeNode] = cloneNode; // Recursively clone the left and right subtrees cloneNode->left = copyLeftRightNode(treeNode->left, mp); cloneNode->right = copyLeftRightNode(treeNode->right, mp); return cloneNode; } // This function copies the random pointers using // the hashmap created in the previous step. void copyRandom(Node *treeNode, unordered_map<Node *, Node *> &mp) { if (treeNode == nullptr) return; // Set the random pointer for the // cloned node mp[treeNode]->random = mp[treeNode->random]; // Recursively copy random pointers for left // and right subtrees copyRandom(treeNode->left, mp); copyRandom(treeNode->right, mp); } // This function creates the clone of // the given binary tree. Node *cloneTree(Node *tree) { if (tree == nullptr) return nullptr; // Create a hashmap to store the mapping // between original and cloned nodes unordered_map<Node *, Node *> mp; // Clone the tree structure // (data, left, and right pointers) Node *newTree = copyLeftRightNode(tree, mp); // Copy the random pointers using // the hashmap copyRandom(tree, mp); return newTree; } // Function to print the inorder traversal of the binary tree // It also prints the random pointer of each node (or NULL if not set) void printInorder(Node *curr) { if (curr == nullptr) return; printInorder(curr->left); cout << "[" << curr->data << " "; if (curr->random == nullptr) cout << "NULL], "; else cout << curr->random->data << "], "; printInorder(curr->right); } int main() { // Constructing the binary tree with random pointers // Tree structure: // 1 // / \ // 2 3 // / \ // 4 5 // Random pointers: // 1 -> 5, 4 -> 1, 5 -> 3 Node *node = new Node(1); node->left = new Node(2); node->right = new Node(3); node->left->left = new Node(4); node->left->right = new Node(5); node->random = node->left->right; node->left->left->random = node; node->left->right->random = node->right; Node *clone = cloneTree(node); printInorder(clone); return 0; } Java // A hashmap-based Java program to clone a binary tree with // random pointers import java.util.HashMap; class Node { int data; Node left, right, random; public Node(int x) { data = x; left = right = random = null; } } class GfG { // This function clones the tree by copying the // key (data) and left/right pointers. static Node copyLeftRightNode(Node treeNode, HashMap<Node, Node> mp) { if (treeNode == null) return null; Node cloneNode = new Node(treeNode.data); // Store the mapping between original node and // cloned node mp.put(treeNode, cloneNode); // Recursively clone the left and right subtrees cloneNode.left = copyLeftRightNode(treeNode.left, mp); cloneNode.right = copyLeftRightNode(treeNode.right, mp); return cloneNode; } // This function copies the random pointers using // the hashmap created in the previous step. static void copyRandom(Node treeNode, HashMap<Node, Node> mp) { if (treeNode == null) return; // Set the random pointer for the cloned node mp.get(treeNode).random = mp.get(treeNode.random); // Recursively copy random pointers for left and // right subtrees copyRandom(treeNode.left, mp); copyRandom(treeNode.right, mp); } // Function to print the inorder traversal of the binary // tree It also prints the random pointer of each node // (or NULL if not set) static void printInorder(Node curr) { if (curr == null) return; printInorder(curr.left); System.out.print("[" + curr.data + " "); if (curr.random == null) System.out.print("NULL], "); else System.out.print(curr.random.data + "], "); printInorder(curr.right); } // This function creates the clone of the given binary // tree. static Node cloneTree(Node tree) { if (tree == null) return null; // Create a hashmap to store the mapping // between original and cloned nodes HashMap<Node, Node> mp = new HashMap<>(); // Clone the tree structure (data, left, and right // pointers) Node newTree = copyLeftRightNode(tree, mp); // Copy the random pointers using the hashmap copyRandom(tree, mp); return newTree; } public static void main(String[] args) { // Constructing the binary tree with random pointers // Tree structure: // 1 // / \ // 2 3 // / \ // 4 5 // Random pointers: // 1 -> 5, 4 -> 1, 5 -> 3 Node node = new Node(1); node.left = new Node(2); node.right = new Node(3); node.left.left = new Node(4); node.left.right = new Node(5); node.random = node.left.right; node.left.left.random = node; node.left.right.random = node.right; Node clone = cloneTree(node); printInorder(clone); } } Python # A hashmap-based Python program to clone a # binary tree with random pointers class Node: def __init__(self, x): self.data = x self.left = self.right = self.random = None # Function to print the inorder traversal of the binary tree # It also prints the random pointer of each node (or NULL if not set) def printInorder(curr): if curr is None: return printInorder(curr.left) print(f"[{curr.data} ", end="") if curr.random is None: print("NULL], ", end="") else: print(f"{curr.random.data}], ", end="") printInorder(curr.right) # This function clones the tree by copying the # key (data) and left/right pointers. def copyLeftRightNode(treeNode, mp): if treeNode is None: return None cloneNode = Node(treeNode.data) # Store the mapping between original # node and cloned node mp[treeNode] = cloneNode # Recursively clone the left and right subtrees cloneNode.left = copyLeftRightNode(treeNode.left, mp) cloneNode.right = copyLeftRightNode(treeNode.right, mp) return cloneNode # This function copies the random pointers using # the hashmap created in the previous step. def copyRandom(treeNode, mp): if treeNode is None: return # Set the random pointer for # the cloned node mp[treeNode].random = mp.get(treeNode.random) # Recursively copy random pointers for # left and right subtrees copyRandom(treeNode.left, mp) copyRandom(treeNode.right, mp) # This function creates the clone of the # given binary tree. def cloneTree(tree): if tree is None: return None # Create a hashmap to store the mapping # between original and cloned nodes mp = {} # Clone the tree structure # (data, left, and right pointers) newTree = copyLeftRightNode(tree, mp) # Copy the random pointers using the hashmap copyRandom(tree, mp) return newTree # Constructing the binary tree with random pointers # Tree structure: # 1 # / \ # 2 3 # / \ # 4 5 # Random pointers: # 1 -> 5, 4 -> 1, 5 -> 3 node = Node(1) node.left = Node(2) node.right = Node(3) node.left.left = Node(4) node.left.right = Node(5) node.random = node.left.right node.left.left.random = node node.left.right.random = node.right clone = cloneTree(node) printInorder(clone) C# // A hashmap-based C# program to clone a binary tree with // random pointers using System; using System.Collections.Generic; class Node { public int data; public Node left, right, random; public Node(int x) { data = x; left = right = random = null; } } class GfG { // Function to print the inorder traversal of the binary // tree It also prints the random pointer of each node // (or NULL if not set) static void PrintInorder(Node curr) { if (curr == null) return; PrintInorder(curr.left); Console.Write("[" + curr.data + " "); if (curr.random == null) Console.Write("NULL], "); else Console.Write(curr.random.data + "], "); PrintInorder(curr.right); } // This function clones the tree by copying the // key (data) and left/right pointers. static Node CopyLeftRightNode(Node treeNode, Dictionary<Node, Node> mp) { if (treeNode == null) return null; Node cloneNode = new Node(treeNode.data); // Store the mapping between original node and // cloned node mp[treeNode] = cloneNode; // Recursively clone the left and right subtrees cloneNode.left = CopyLeftRightNode(treeNode.left, mp); cloneNode.right = CopyLeftRightNode(treeNode.right, mp); return cloneNode; } // This function copies the random pointers using // the hashmap created in the previous step. static void CopyRandom(Node treeNode, Dictionary<Node, Node> mp) { if (treeNode == null) return; // Set the random pointer for the // cloned node if (treeNode.random != null) { mp[treeNode].random = mp[treeNode.random]; } // Recursively copy random pointers for left and // right subtrees CopyRandom(treeNode.left, mp); CopyRandom(treeNode.right, mp); } // This function creates the clone of the given binary // tree. static Node CloneTree(Node tree) { if (tree == null) return null; // Create a hashmap to store the mapping // between original and cloned nodes Dictionary<Node, Node> mp = new Dictionary<Node, Node>(); // Clone the tree structure (data, left, and right // pointers) Node newTree = CopyLeftRightNode(tree, mp); // Copy the random pointers using the hashmap CopyRandom(tree, mp); return newTree; } static void Main(string[] args) { // Constructing the binary tree with random pointers // Tree structure: // 1 // / \ // 2 3 // / \ // 4 5 // Random pointers: // 1 -> 5, 4 -> 1, 5 -> 3 Node node = new Node(1); node.left = new Node(2); node.right = new Node(3); node.left.left = new Node(4); node.left.right = new Node(5); node.random = node.left.right; node.left.left.random = node; node.left.right.random = node.right; Node clone = CloneTree(node); PrintInorder(clone); } } JavaScript // A hashmap-based JavaScript program to clone a binary tree // with random pointers class Node { constructor(x) { this.data = x; this.left = this.right = this.random = null; } } // Function to print the inorder traversal of the binary // tree It also prints the random pointer of each node (or // NULL if not set) function printInorder(curr) { if (curr === null) return; printInorder(curr.left); console.log(`[${curr.data} `); if (curr.random === undefined || curr.random === null) console.log("NULL], "); else console.log(`${curr.random.data}], `); printInorder(curr.right); } // This function clones the tree by copying the // key (data) and left/right pointers. function copyLeftRightNode(treeNode, mp) { if (treeNode === null) return null; let cloneNode = new Node(treeNode.data); // Store the mapping between original node and cloned // node mp.set(treeNode, cloneNode); // Recursively clone the left and right subtrees cloneNode.left = copyLeftRightNode(treeNode.left, mp); cloneNode.right = copyLeftRightNode(treeNode.right, mp); return cloneNode; } // This function copies the random pointers using // the hashmap created in the previous step. function copyRandom(treeNode, mp) { if (treeNode === null) return; // Set the random pointer for the cloned node mp.get(treeNode).random = mp.get(treeNode.random); // Recursively copy random pointers for left and right // subtrees copyRandom(treeNode.left, mp); copyRandom(treeNode.right, mp); } // This function creates the clone of the given binary tree. function cloneTree(tree) { if (tree === null) return null; // Create a hashmap to store the mapping // between original and cloned nodes let mp = new Map(); // Clone the tree structure (data, left, and right // pointers) let newTree = copyLeftRightNode(tree, mp); // Copy the random pointers using the hashmap copyRandom(tree, mp); return newTree; } // Constructing the binary tree with random pointers // Tree structure: // 1 // / \ // 2 3 // / \ // 4 5 // Random pointers: // 1 -> 5, 4 -> 1, 5 -> 3 let node = new Node(1); node.left = new Node(2); node.right = new Node(3); node.left.left = new Node(4); node.left.right = new Node(5); node.random = node.left.right; node.left.left.random = node; node.left.right.random = node.right; let clone = cloneTree(node); printInorder(clone); Output[4 1], [2 NULL], [5 3], [1 5], [3 NULL], Time complexity: O(n), this is because we need to traverse the entire tree in order to copy the left and right pointers, and then we need to traverse the tree again to copy the random pointers.Auxiliary Space: O(n), this is because we need to store a mapping of the original tree’s nodes to their clones. Comment More infoAdvertise with us Next Article Clone a Binary Tree with Random Pointers kartik Follow Improve Article Tags : Tree Hash DSA Amazon Practice Tags : AmazonHashTree Similar Reads Hashing in Data Structure Hashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The 3 min read Introduction to Hashing Hashing refers to the process of generating a small sized output (that can be used as index in a table) from an input of typically large and variable size. Hashing uses mathematical formulas known as hash functions to do the transformation. This technique determines an index or location for the stor 7 min read What is Hashing? 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Examples: Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12} Output : 6 Explanation: 3, 4, 5, 6, 7, 8 is the longest increasing subsequence whose adjacent element differs by one. Input : a[] = {6 10 min read Count Distinct Elements In Every Window of Size KGiven an array arr[] of size n and an integer k, return the count of distinct numbers in all windows of size k. Examples: Input: arr[] = [1, 2, 1, 3, 4, 2, 3], k = 4Output: [3, 4, 4, 3]Explanation: First window is [1, 2, 1, 3], count of distinct numbers is 3. Second window is [2, 1, 3, 4] count of d 10 min read Design a data structure that supports insert, delete, search and getRandom in constant timeDesign a data structure that supports the following operations in O(1) time.insert(x): Inserts an item x to the data structure if not already present.remove(x): Removes item x from the data structure if present. search(x): Searches an item x in the data structure.getRandom(): Returns a random elemen 5 min read Subarray with Given Sum - Handles Negative NumbersGiven an unsorted array of integers, find a subarray that adds to a given number. 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Once an empty slot is 13 min read Maximum possible difference of two subsets of an arrayGiven an array of n-integers. The array may contain repetitive elements but the highest frequency of any element must not exceed two. You have to make two subsets such that the difference of the sum of their elements is maximum and both of them jointly contain all elements of the given array along w 15+ min read Sorting using trivial hash functionWe have read about various sorting algorithms such as heap sort, bubble sort, merge sort and others. Here we will see how can we sort N elements using a hash array. But this algorithm has a limitation. We can sort only those N elements, where the value of elements is not large (typically not above 1 15+ min read Smallest subarray with k distinct numbersWe are given an array consisting of n integers and an integer k. We need to find the smallest subarray [l, r] (both l and r are inclusive) such that there are exactly k different numbers. If no such subarray exists, print -1 and If multiple subarrays meet the criteria, return the one with the smalle 14 min read Like