Maximize sum of path from the Root to a Leaf node in N-ary Tree Last Updated : 23 Nov, 2024 Comments Improve Suggest changes Like Article Like Report Given a generic tree consisting of n nodes, the task is to find the maximum sum of the path from the root to the leaf node.Examples:Input:Output: 12Explanation: The path sum to every leaf from the root are:For node 4: 1 -> 2 -> 4 = 7For node 5: 1 -> 2 -> 5 = 8For node 6: 1 -> 3 -> 6 = 10For node 7: 1 -> 3 -> 7 = 11For node 8: 1 -> 3 -> 8 = 12 (maximum)The maximum path sum is 12 i.e., from root 1 to leaf 8.Approach:The given problem can be solved by performing the DFS traversal on the given tree. The idea is to perform the DFS Traversal from the root node of the given generic tree by keeping the track of the sum of values of nodes in each path and if any leaf node occurs then maximize the value of the current sum of path obtained in a variable, say maxSum.After performing the DFS Traversal, print the value of maxSum as the resultant maximum sum path. C++ // C++ code of finding Maximize sum of path from // the Root to a Leaf node in N-ary Tree #include <bits/stdc++.h> using namespace std; struct Node { int data; vector<Node*> child; Node(int x){ data = x; } }; // Recursive function to calculate the // maximum sum in a path using DFS void DFS(Node* root, int sum, int& ans) { // If current node is a leaf node if (root->child.size() == 0) { ans = max(ans, sum); return; } // Traversing all children of // the current node for (int i = 0; i < root->child.size(); i++) { // Recursive call for all // the children nodes DFS(root->child[i], sum + root->child[i]->data, ans); } } int main() { Node* root = new Node(1); (root->child).push_back(new Node(2)); (root->child).push_back(new Node(3)); (root->child[0]->child).push_back(new Node(4)); (root->child[1]->child).push_back(new Node(6)); (root->child[0]->child).push_back(new Node(5)); (root->child[1])->child.push_back(new Node(7)); (root->child[1]->child).push_back(new Node(8)); int maxSumPath = 0; DFS(root, root->data, maxSumPath); cout << maxSumPath; return 0; } Java // Java code to find the maximum sum of a path from // the Root to a Leaf node in an N-ary Tree import java.util.ArrayList; class Node { int data; ArrayList<Node> child; Node(int x) { data = x; child = new ArrayList<>(); } } class GfG { // Recursive function to calculate the // maximum sum in a path using DFS static void DFS(Node root, int sum, int[] ans) { // If the current node is a leaf node if (root.child.size() == 0) { ans[0] = Math.max(ans[0], sum); return; } // Traversing all children of // the current node for (int i = 0; i < root.child.size(); i++) { // Recursive call for all // the children nodes DFS(root.child.get(i), sum + root.child.get(i).data, ans); } } public static void main(String[] args) { Node root = new Node(1); root.child.add(new Node(2)); root.child.add(new Node(3)); root.child.get(0).child.add(new Node(4)); root.child.get(1).child.add(new Node(6)); root.child.get(0).child.add(new Node(5)); root.child.get(1).child.add(new Node(7)); root.child.get(1).child.add(new Node(8)); int[] maxSumPath = new int[1]; maxSumPath[0] = 0; DFS(root, root.data, maxSumPath); System.out.println(maxSumPath[0]); } } Python # Python code to find the maximum sum of a path from # the Root to a Leaf node in an N-ary Tree class Node: def __init__(self, x): self.data = x self.child = [] # Recursive function to calculate the # maximum sum in a path using DFS def DFS(root, sum, ans): # If the current node is a leaf node if len(root.child) == 0: ans[0] = max(ans[0], sum) return # Traversing all children of # the current node for child in root.child: # Recursive call for all # the children nodes DFS(child, sum + child.data, ans) if __name__ == "__main__": root = Node(1) root.child.append(Node(2)) root.child.append(Node(3)) root.child[0].child.append(Node(4)) root.child[1].child.append(Node(6)) root.child[0].child.append(Node(5)) root.child[1].child.append(Node(7)) root.child[1].child.append(Node(8)) maxSumPath = [0] DFS(root, root.data, maxSumPath) print(maxSumPath[0]) C# // C# code to find the maximum sum of a path from // the Root to a Leaf node in an N-ary Tree using System; using System.Collections.Generic; class Node { public int data; public List<Node> child; public Node(int x) { data = x; child = new List<Node>(); } } class GfG { // Recursive function to calculate the // maximum sum in a path using DFS static void DFS(Node root, int sum, ref int ans) { // If the current node is a leaf node if (root.child.Count == 0) { ans = Math.Max(ans, sum); return; } // Traversing all children of // the current node foreach (var child in root.child) { // Recursive call for all // the children nodes DFS(child, sum + child.data, ref ans); } } static void Main() { Node root = new Node(1); root.child.Add(new Node(2)); root.child.Add(new Node(3)); root.child[0].child.Add(new Node(4)); root.child[1].child.Add(new Node(6)); root.child[0].child.Add(new Node(5)); root.child[1].child.Add(new Node(7)); root.child[1].child.Add(new Node(8)); int maxSumPath = 0; DFS(root, root.data, ref maxSumPath); Console.WriteLine(maxSumPath); } } JavaScript // JavaScript code to find the maximum sum of a path from // the Root to a Leaf node in an N-ary Tree class Node { constructor(x) { this.data = x; this.child = []; } } // Recursive function to calculate the // maximum sum in a path using DFS function DFS(root, sum, ans) { // If the current node is a leaf node if (root.child.length === 0) { ans[0] = Math.max(ans[0], sum); return; } // Traversing all children of // the current node for (let i = 0; i < root.child.length; i++) { // Recursive call for all // the children nodes DFS(root.child[i], sum + root.child[i].data, ans); } } let root = new Node(1); root.child.push(new Node(2)); root.child.push(new Node(3)); root.child[0].child.push(new Node(4)); root.child[1].child.push(new Node(6)); root.child[0].child.push(new Node(5)); root.child[1].child.push(new Node(7)); root.child[1].child.push(new Node(8)); let maxSumPath = [0]; DFS(root, root.data, maxSumPath); console.log(maxSumPath[0]); Output12Time Complexity: O(n), where n is the number of nodes in treeAuxiliary Space: O(n) Comment More infoAdvertise with us Next Article Maximize sum of path from the Root to a Leaf node in N-ary Tree S sohailahmed46khan786 Follow Improve Article Tags : Misc Tree Mathematical Recursion DSA DFS Tree Traversals n-ary-tree +4 More Practice Tags : DFSMathematicalMiscRecursionTree +1 More Similar Reads Introduction to Generic Trees (N-ary Trees) Generic trees are a collection of nodes where each node is a data structure that consists of records and a list of references to its children(duplicate references are not allowed). 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Examples Input: 1 / | \ 2 3 4 / \ \ 5 6 7Output: Initial Pre-order Traversal: 1 2 5 6 7 3 4 Final Pre-order Traversal: 28 20 5 6 7 3 4 Explanation: Value of each node is replaced 8 min read Path from the root node to a given node in an N-ary TreeGiven an integer N and an N-ary Tree of the following form: Every node is numbered sequentially, starting from 1, till the last level, which contains the node N.The nodes at every odd level contains 2 children and nodes at every even level contains 4 children. The task is to print the path from the 10 min read Determine the count of Leaf nodes in an N-ary treeGiven the value of 'N' and 'I'. Here, I represents the number of internal nodes present in an N-ary tree and every node of the N-ary can either have N childs or zero child. The task is to determine the number of Leaf nodes in n-ary tree. Examples: Input : N = 3, I = 5 Output : Leaf nodes = 11 Input 4 min read Remove all leaf nodes from a Generic Tree or N-ary TreeGiven an n-ary tree containing positive node values, the task is to delete all the leaf nodes from the tree and print preorder traversal of the tree after performing the deletion.Note: An n-ary tree is a tree where each node can have zero or more children nodes. Unlike a binary tree, which has at mo 6 min read Maximum level sum in N-ary TreeGiven an N-ary Tree consisting of nodes valued [1, N] and an array value[], where each node i is associated with value[i], the task is to find the maximum of sum of all node values of all levels of the N-ary Tree. Examples: Input: N = 8, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, 9 min read Number of leaf nodes in a perfect N-ary tree of height KFind the number of leaf nodes in a perfect N-ary tree of height K. Note: As the answer can be very large, return the answer modulo 109+7. Examples: Input: N = 2, K = 2Output: 4Explanation: A perfect Binary tree of height 2 has 4 leaf nodes. Input: N = 2, K = 1Output: 2Explanation: A perfect Binary t 4 min read Print all root to leaf paths of an N-ary treeGiven an N-Ary tree, the task is to print all root to leaf paths of the given N-ary Tree. Examples: Input: 1 / \ 2 3 / / \ 4 5 6 / \ 7 8 Output:1 2 41 3 51 3 6 71 3 6 8 Input: 1 / | \ 2 5 3 / \ \ 4 5 6Output:1 2 41 2 51 51 3 6 Approach: The idea to solve this problem is to start traversing the N-ary 7 min read Minimum distance between two given nodes in an N-ary treeGiven a N ary Tree consisting of N nodes, the task is to find the minimum distance from node A to node B of the tree. Examples: Input: 1 / \ 2 3 / \ / \ \4 5 6 7 8A = 4, B = 3Output: 3Explanation: The path 4->2->1->3 gives the minimum distance from A to B. 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