N-ary Tree Level Order Traversal - Easy BFS solution

Breadth-First Search     Medium     Tree    

Problem Statement:

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]

Solution:

We had this solution for Level order traversal in Binary Tree:

void helper(TreeNode *root, vector<vector<int>> &res, int level)
{
	if (!root) return;
	if (res.size()==level) res.push_back({});
	res[level].push_back(root->val);
	helper(root->left, res, level+1);
	helper(root->right, res, level+1);
}
vector<vector<int>> levelOrder(TreeNode* root) 
{
	vector<vector<int>> res;
	helper(root, res, 0);
	return res;
}

We can modify this for N-ary tree as:

void helper(Node *root, vector<vector<int>> &res, int level)
{
	if (!root) return;
	if (res.size()==level) res.push_back({});
	res[level].push_back(root->val);
	for (auto child: root->children) helper(child, res, level+1);
}
vector<vector<int>> levelOrder(Node* root) 
{
	vector<vector<int>> res;
	helper(root, res, 0);
	return res;
}