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Formatted question description: https://leetcode.ca/all/429.html

# 429. N-ary Tree Level Order Traversal (Easy)

Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example, given a 3-ary tree:

We should return its level order traversal:

[
[1],
[3,2,4],
[5,6]
]

Note:

1. The depth of the tree is at most 1000.
2. The total number of nodes is at most 5000.

## Solution 1.

• /*
// 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>> levelOrder = new ArrayList<List<Integer>>();
if (root == null)
return levelOrder;
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> curLevel = new ArrayList<Integer>();
for (int i = 0; i < size; i++) {
Node node = queue.poll();
List<Node> children = node.children;
if (children != null && children.size() > 0) {
int childrenCount = children.size();
for (int j = 0; j < childrenCount; j++)
queue.offer(children.get(j));
}
}
}
return levelOrder;
}
}

############

/*
// 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;
}
}

• // OJ: https://leetcode.com/problems/n-ary-tree-level-order-traversal/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<vector<int>> levelOrder(Node* root) {
if (!root) return {};
vector<vector<int>> ans;
queue<Node*> q{ {root} };
while (q.size()) {
int cnt = q.size();
vector<int> level;
while (cnt--) {
auto node = q.front();
q.pop();
level.push_back(node->val);
for (auto &child : node->children) q.push(child);
}
ans.push_back(level);
}
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.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def levelOrder(self, root):
"""
:type root: Node
:rtype: List[List[int]]
"""
if not root:
return []
queue = [(root, 0)]
res = [[]]
while queue:
node, level = queue.pop(0)
if level >= len(res):
res.append([])
res[level].append(node.val)
for child in node.children:
queue.append((child, level + 1))
return res

• /**
* 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;
}