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# N-ary Tree Level Order Traversal LeetCode Solution

## Problem – N-ary Tree Level Order Traversal LeetCode Solution

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:

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

Example 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: [,[2,3,4,5],[6,7,8,9,10],[11,12,13],]
``````

Constraints:

• The height of the n-ary tree is less than or equal to `1000`
• The total number of nodes is between `[0, 104]`

### N-ary Tree Level Order Traversal LeetCode Solution in Python

``````class Solution(object):
def levelOrder(self, root):
q, ret = [root], []
while any(q):
ret.append([node.val for node in q])
q = [child for node in q for child in node.children if child]
return ret
``````

### N-ary Tree Level Order Traversal LeetCode Solution in Java

``````public List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> ret = new LinkedList<>();

if (root == null) return ret;

Queue<Node> queue = new LinkedList<>();

queue.offer(root);

while (!queue.isEmpty()) {
List<Integer> curLevel = new LinkedList<>();
int len = queue.size();
for (int i = 0; i < len; i++) {
Node curr = queue.poll();
for (Node c : curr.children)
queue.offer(c);
}
}

return ret;
}
``````

### N-ary Tree Level Order Traversal LeetCode Solution in C++

``````class Solution {
public:
vector<vector<int>> levelOrder(Node* root) {
if (root == NULL)
return {};
vector<vector<int>> res;
queue<Node*> q;
q.push(root);
while (!q.empty())
{
int size = q.size();
vector<int> currlevel;
for(int i=0;i<size;i++){
Node* tmp=q.front();
q.pop();
currlevel.push_back(tmp->val);
for (auto n : tmp -> children)
q.push(n);
}
res.push_back(currlevel);
}
return res;
}
};
``````
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