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.

// 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 {};
        queue<Node*> q;
        q.push(root);
        vector<vector<int>> ans;
        while (q.size()) {
            int cnt = q.size();
            vector<int> level;
            while (cnt--) {
                root = q.front();
                q.pop();
                level.push_back(root->val);
                for (Node *node : root->children) q.push(node);
            }
            ans.push_back(level);
        }
        return ans;
    }
};

Java

  • /*
    // 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<Node> queue = new LinkedList<Node>();
            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();
                    curLevel.add(node.val);
                    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));
                    }
                }
                levelOrder.add(curLevel);
            }
            return levelOrder;
        }
    }
    
  • // 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(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
    

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