Find the Root of a Tree Last Updated : 22 Jan, 2024 Comments Improve Suggest changes Like Article Like Report Given a random node of a tree, find out the root of the tree. Note: Every node has a value (val) and a pointer (*parent) pointing towards its parent. Examples: Input: A Node : E / \ B D / | / \ E G H K Output: A Explanation: A is the root of the tree Input: 1 Node : 6 / \ 2 3 / \ / 4 5 6 Output: 1 Explanation: 1 is the root of the tree Approach: To solve the problem follow the below idea: Traverse the tree: Start from any node in the tree and traverse upward by following parent pointers until you reach a node with no parent (i.e., the root).Check node parent: In each step of the traversal, check if the current node has a parent. If it does, move to the parent node and continue the traversal.Stop at the root: Repeat the traversal until you reach the root node (the node with no parent). C++ #include<iostream>; #include<unordered_map>; using namespace std; struct TreeNode { int val; TreeNode* parent; TreeNode(int value, TreeNode* p = nullptr) : val(value) , parent(p) { } }; // Function to find the root of the tree TreeNode* findRoot(TreeNode* node) { // Traverse upward until we reach a node with no parent while (node!=NULL && node->parent) { node = node->parent; } return node; } int main() { // Create a tree TreeNode* root = new TreeNode(1); TreeNode* node2 = new TreeNode(2, root); TreeNode* node3 = new TreeNode(3, root); TreeNode* node4 = new TreeNode(4, node2); TreeNode* node5 = new TreeNode(5, node2); TreeNode* node6 = new TreeNode(6, node3); // Find the root of the tree TreeNode* treeRoot = findRoot(node6); cout<<"Root value: "<<treeRoot->val<<endl; // Clean up memory delete node6; delete node5; delete node4; delete node3; delete node2; delete root; return 0; } Java /*package whatever //do not write package name here */ import java.io.*; class TreeNode { int val; TreeNode parent; public TreeNode(int value, TreeNode p) { val = value; parent = p; } } public class Main { public static TreeNode findRoot(TreeNode node) { // Traverse upward until we reach a node with no parent while (node != null && node.parent != null) { node = node.parent; } return node; } public static void main(String[] args) { // Create a tree TreeNode root = new TreeNode(1, null); TreeNode node2 = new TreeNode(2, root); TreeNode node3 = new TreeNode(3, root); TreeNode node4 = new TreeNode(4, node2); TreeNode node5 = new TreeNode(5, node2); TreeNode node6 = new TreeNode(6, node3); // Find the root of the tree TreeNode treeRoot = findRoot(node6); System.out.println("Root value: " + treeRoot.val); } } Python3 class TreeNode: def __init__(self, value, parent=None): self.val = value self.parent = parent # Function to find the root of the tree def find_root(node): # Traverse upward until we reach a node with no parent while node is not None and node.parent is not None: node = node.parent return node def main(): # Create a tree root = TreeNode(1) node2 = TreeNode(2, root) node3 = TreeNode(3, root) node4 = TreeNode(4, node2) node5 = TreeNode(5, node2) node6 = TreeNode(6, node3) # Find the root of the tree tree_root = find_root(node6) print("Root value:", tree_root.val) # Clean up memory (optional in Python) node6 = node5 = node4 = node3 = node2 = root = None if __name__ == "__main__": main() C# using System; // TreeNode class representing a node in the tree public class TreeNode { public int Val { get; } public TreeNode Parent { get; set; } // Constructor to initialize a TreeNode with a value and optional parent public TreeNode(int value, TreeNode parent = null) { Val = value; Parent = parent; } } class Program { // Function to find the root of the tree static TreeNode FindRoot(TreeNode node) { // Traverse upward until we reach a node with no parent while (node != null && node.Parent != null) { node = node.Parent; } return node; } static void Main() { // Create a tree TreeNode root = new TreeNode(1); // Removed node2 since it is not used // TreeNode node2 = new TreeNode(2, root); TreeNode node3 = new TreeNode(3, root); TreeNode node6 = new TreeNode(6, node3); // Find the root of the tree TreeNode treeRoot = FindRoot(node6); Console.WriteLine($"Root value: {treeRoot.Val}"); // Clean up memory (assuming garbage collection in C#) // Note: In C#, explicit memory cleanup is often not needed due to automatic garbage collection. // You don't need to delete objects as you do in C++. // If you want to release resources explicitly, you can use IDisposable and implement the Dispose method. // For simplicity, explicit deletion is omitted in this example. // Note: C# provides automatic memory management, so there's usually no need to manually delete objects. } } JavaScript class TreeNode { constructor(value, parent = null) { this.val = value; this.parent = parent; } } // Function to find the root of the tree function GFG(node) { // Traverse upward until we reach a node with // no parent while (node !== null && node.parent !== null) { node = node.parent; } return node; } // Main function function main() { // Create a tree const root = new TreeNode(1); const node2 = new TreeNode(2, root); const node3 = new TreeNode(3, root); const node4 = new TreeNode(4, node2); const node5 = new TreeNode(5, node2); const node6 = new TreeNode(6, node3); // Find the root of the tree const treeRoot = GFG(node6); // Output the root value console.log("Root value:", treeRoot.val); } main(); OutputRoot value: 1 Time Complexity: O(N), where N is the number of nodesAuxiliary Space: O(1) Comment More infoAdvertise with us N nikhilgarg527 Follow Improve Article Tags : Tree DSA Practice Tags : Tree 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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