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.

/*
// 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;
    }
}

############

/*
// 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>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                root = q.poll();
                t.add(root.val);
                q.addAll(root.children);
            }
            ans.add(t);
        }
        return ans;
    }
}

// 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:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def levelOrder(self, root: 'Node') -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                root = q.popleft()
                t.append(root.val)
                q.extend(root.children)
            ans.append(t)
        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

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func levelOrder(root *Node) [][]int {
	var ans [][]int
	if root == nil {
		return ans
	}
	q := []*Node{root}
	for len(q) > 0 {
		var t []int
		for n := len(q); n > 0; n-- {
			root = q[0]
			q = q[1:]
			t = append(t, root.Val)
			for _, child := range root.Children {
				q = append(q, child)
			}
		}
		ans = append(ans, t)
	}
	return ans
}

/**
 * Definition for node.
 * class Node {
 *     val: number
 *     children: Node[]
 *     constructor(val?: number) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.children = []
 *     }
 * }
 */

function levelOrder(root: Node | null): number[][] {
    const res = [];
    if (root == null) {
        return res;
    }
    const queue = [root];
    while (queue.length !== 0) {
        const n = queue.length;
        const vals = [];
        for (let i = 0; i < n; i++) {
            const { val, children } = queue.shift();
            vals.push(val);
            queue.push(...children);
        }
        res.push(vals);
    }
    return res;
}

429 - N-ary Tree Level Order Traversal

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

Solution 1.

All Problems

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

Solution 1.

All Problems

All Solutions