Algorithm
Problem Name: 429. N-ary Tree Level Order Traversal
Problem Link: https://leetcode.com/problems/n-ary-tree-level-order-traversal/
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]
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public List();
}
List();
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
List < Integer> currLevel = new ArrayList<>();
while (size-- > 0) {
Node removed = queue.remove();
currLevel.add(removed.val);
queue.addAll(removed.children);
}
levels.add(currLevel);
}
return levels;
}
}
Input
root = [1,null,3,2,4,null,5,6]
Output
[[1],[3,2,4],[5,6]]
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const levelOrder = function(root) {
const res = []
if(root == null) return res
helper(root, 0, res)
return res
};
function helper(node, index, res) {
if(node == null) return
if(res[index] == null) res[index] = []
res[index].push(node.val)
for(let i = 0, len = node.children.length; i < len; i++) {
helper(node.children[i], index + 1, res>
}
}
Input
root = [1,null,3,2,4,null,5,6]
Output
[[1],[3,2,4],[5,6]]
#3 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0429_NAryTreeLevelOrderTraversal
{
public IList < IList<int>> LevelOrder(Node root)
{
var results = new List < IList<int>>();
if (root == null) return results;
var queue = new Queue < Node>();
queue.Enqueue(root);
while (queue.Count > 0)
{
var size = queue.Count;
var result = new List < int>();
for (int i = 0; i < size; i++)
{
var node = queue.Dequeue();
result.Add(node.val);
if (node.children != null)
foreach (var child in node.children)
queue.Enqueue(child);
}
results.Add(result);
}
return results;
}
public class Node
{
public int val;
public IList < Node> children;
public Node() { }
public Node(int _val)
{
val = _val;
}
public Node(int _val, IList < Node> _children)
{
val = _val;
children = _children;
}
}
}
}
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]]