429. N-ary Tree Level Order Traversal

2020, Nov 22 2 mins read

1. 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).

2. Example

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]]

3. Constraints

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10$^{4}$]

4. Solutions

Breadth-First Search

n is the number of nodes in i_root
Time complexity: O(n)
Space complexity: O(n)

class Solution
{
public:
    // vector<vector<int>> levelOrder(Node* root)
    vector<vector<int>> levelOrder(Node *i_root)
    {
        if(!i_root) return vector<vector<int>>{};

        vector<vector<int>> result;
        queue<Node *> currentLevelNodes{{i_root}};
        queue<Node *> nextLevelNodes;

        vector<int> levelNodesValue;
        while(!currentLevelNodes.empty() || !nextLevelNodes.empty())
        {
            if(currentLevelNodes.empty())
            {
                result.push_back(levelNodesValue);
                levelNodesValue.clear();

                swap(currentLevelNodes, nextLevelNodes);
            }
            else
            {
                auto node = currentLevelNodes.front();
                currentLevelNodes.pop();

                levelNodesValue.push_back(node->val);

                for(auto iter : node->children)
                {
                    if(iter) nextLevelNodes.push(iter);
                }
            }
        }

        result.push_back(levelNodesValue);

        return result;
    }
};

Recursion

n is the number of nodes in i_root
Time complexity: O(n)
Space complexity: O(n)

class Solution
{
private:
    vector<vector<int>> result;

    void traverseNode(Node *i_root, int level)
    {
        if(!i_root) return;

        if(result.size() <= level) result.push_back(vector<int>{});

        result[level].push_back(i_root->val);
        for(auto child : i_root->children)
        {
            traverseNode(child, level + 1);
        }
    }
public:
    // vector<vector<int>> levelOrder(Node* root)
    vector<vector<int>> levelOrder(Node *i_root)
    {
        traverseNode(i_root, 0);

        return result;
    }
};