Leetcode N-ary Tree Level Order Traversal problem solution YASH PAL, 31 July 2024 In this Leetcode N-ary Tree Level Order Traversal problem solution we have given an n-ary tree, returns the level order traversal of its nodes’ values. Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++. Problem solution in Python. class Solution(object): def levelOrder(self, root): if not root:return [] res = [] stack = [root] while stack: temp = [] next_stack = [] for node in stack: temp.append(node.val) for child in node.children: next_stack.append(child) stack = next_stack res.append(temp) return res Problem solution in Java. class Solution { public List<List> levelOrder(Node root) { Queue <Node> q1=new LinkedList(); Queue <Node> q2=new LinkedList(); List<List<Integer>> list=new LinkedList<List<Integer>>(); if(root==null){ return list; } List<Integer> list1=new ArrayList<>(); q1.add(root); while(q1.size()!=0){ Node a =q1.remove(); // System.out.print(a.val+" "); list1.add(a.val); for(Node child:a.children){ q2.add(child); } if(q1.size()==0){ q1=q2; q2=new LinkedList(); list.add(list1); list1=new ArrayList<>(); // System.out.println(); } } return list; } } Problem solution in C++. class Solution { public: vector<vector<int>> levelOrder(Node* root) { vector<vector<int>> ret; if(root == nullptr) return ret; queue<Node*> s; queue<int> num; s.push(root); num.push(1); while(!s.empty()) { int N = num.front(); num.pop(); int Nnext = 0; ret.push_back(vector<int>()); for(int n=0; n<N; ++n) { Node *node = s.front(); s.pop(); Nnext += node->children.size(); ret.back().push_back(node->val); for(int k=0; k<node->children.size(); ++k) s.push(node->children[k]); } num.push(Nnext); } return ret; } }; coding problems solutions