429

429. N-ary Tree Level Order Traversal (Medium)

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]

Companies: Amazon, Adobe, Google

Related Topics:
Tree, Breadth-First Search

Similar Questions:

• Binary Tree Level Order Traversal (Medium)
• N-ary Tree Preorder Traversal (Easy)
• N-ary Tree Postorder Traversal (Easy)
• The Time When the Network Becomes Idle (Medium)

Solution 1. BFS

// OJ: https://leetcode.com/problems/n-ary-tree-level-order-traversal/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/n-ary-tree-level-order-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    vector<vector<int>> ans;
    void dfs(Node *root, int level) {
        if (!root) return;
        if (level >= ans.size()) ans.emplace_back();
        ans[level].push_back(root->val);
        for (auto &child : root->children) dfs(child, level + 1);
    }
public:
    vector<vector<int>> levelOrder(Node* root) {
        dfs(root, 0);
        return ans;
    }
};