429. N-ary Tree Level Order Traversal

MediumView on LeetCode

Tree Breadth-First Search

Time: O(n)

Space: O(n)

Intuition

BFS level-by-level, same as binary tree but each node can have multiple children.

Algorithm

1Start BFS queue with root.
2For each level: record count of nodes in queue, dequeue that many nodes collecting their values, enqueue all their children.
3Append each level's values to result.

Example Walkthrough

Input: root=[1,[3,2,4],[5,6]]

1.Level 0: [1].
2.Level 1: [3,2,4].
3.Level 2: [5,6].

Output: [[1],[3,2,4],[5,6]]

Common Pitfalls

•Snapshot the queue size before processing each level, otherwise newly added children corrupt level count.

429.cs

// Approach: BFS with a queue; collect all children of dequeued nodes
// at each level into the next level’s snapshot.
// Time: O(n) Space: O(n)

public class Node
{
    public int val;
    public IList<Node> children;

    public Node() { }

    public Node(int _val)
    {
        val = _val;
    }

    public Node(int _val, IList<Node> _children)
    {
        val = _val;
        children = _children;
    }
}


public class Solution
{
    public IList<IList<int>> LevelOrder(Node root)
    {
        if (root == null)
            return new List<IList<int>>();

        IList<IList<int>> ans = new List<IList<int>>();
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);

        while (q.Count > 0)
        {
            IList<int> currLevel = new List<int>();
            int sz = q.Count;
            for (int i = 0; i < sz; i++)
            {
                Node node = q.Dequeue();
                currLevel.Add(node.val);
                foreach (Node child in node.children)
                    q.Enqueue(child);
            }
            ans.Add(currLevel);
        }

        return ans;
    }
}

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