# 429. N-ary Tree Level Order Traversal

## Description

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: [[1],[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: [[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, 104]

## Solutions

BFS or DFS.

• /*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;

public Node() {}

public Node(int _val) {
val = _val;
}

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

class Solution {
public List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<Node> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
List<Integer> t = new ArrayList<>();
for (int n = q.size(); n > 0; --n) {
root = q.poll();
}
}
return ans;
}
}

• /*
// 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) {
vector<vector<int>> ans;
if (!root) return ans;
queue<Node*> q{ {root} };
while (!q.empty()) {
vector<int> t;
for (int n = q.size(); n > 0; --n) {
root = q.front();
q.pop();
t.push_back(root->val);
for (auto& child : root->children) q.push(child);
}
ans.push_back(t);
}
return ans;
}
};

• """
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""

class Solution:
def levelOrder(self, root: 'Node') -> List[List[int]]:
ans = []
if root is None:
return ans
q = deque([root])
while q:
t = []
for _ in range(len(q)):
root = q.popleft()
t.append(root.val)
q.extend(root.children)
ans.append(t)
return ans


• /**
* Definition for a Node.
* type Node struct {
*     Val int
*     Children []*Node
* }
*/

func levelOrder(root *Node) [][]int {
var ans [][]int
if root == nil {
return ans
}
q := []*Node{root}
for len(q) > 0 {
var t []int
for n := len(q); n > 0; n-- {
root = q[0]
q = q[1:]
t = append(t, root.Val)
for _, child := range root.Children {
q = append(q, child)
}
}
ans = append(ans, t)
}
return ans
}

• /**
* Definition for node.
* class Node {
*     val: number
*     children: Node[]
*     constructor(val?: number) {
*         this.val = (val===undefined ? 0 : val)
*         this.children = []
*     }
* }
*/

function levelOrder(root: Node | null): number[][] {
const res = [];
if (root == null) {
return res;
}
const queue = [root];
while (queue.length !== 0) {
const n = queue.length;
const vals = [];
for (let i = 0; i < n; i++) {
const { val, children } = queue.shift();
vals.push(val);
queue.push(...children);
}
res.push(vals);
}
return res;
}