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:

The depth of the tree is at most 1000.
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

429 - N-ary Tree Level Order Traversal

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

Solution 1.

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