# N-Ary Tree Level Order Traversal Problem

## Description

LeetCode Problem 429.

Given an n-ary tree, return 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 (See examples).

Example 1:

``````1
2
Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]
``````

Example 2:

``````1
2
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
``````

Constraints:

• The height of the n-ary tree is less than or equal to 1000
• The total number of nodes is between [0, 10^4]

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
/*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children;

Node() {}

Node(int _val) {
val = _val;
}

Node(int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
*/

class Solution {
public:
vector<vector<int>> levelOrder(Node* root) {
if (root == nullptr) return {};

queue<Node*> q;
q.push(root);
vector<vector<int>> ans;

while (!q.empty()) {
ans.emplace_back();
for (int i = q.size(); i >= 1; i--) {
Node* curr = q.front(); q.pop();
ans.back().push_back(curr->val);
for (Node* child : curr->children) {
q.push(child);
}
}
}

return ans;
}
};
``````