Reference: LeetCode
Difficulty: Easy

Problem

Given an n-ary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

Example: 

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[
     [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.

Analysis

Recursion

The recursion is actually based on the preorder traversal.

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public List<List<Integer>> levelOrder(Node root) {
  List<List<Integer>> result = new ArrayList<>();
  preorder(root, 0, result);
  return result;
}

private void preorder(Node root, List<List<Integer>> result, int level) {
  if (root == null) {
    return;
  }
  // expand
  if (result.size() == level) result.add(new ArrayList<>());
  // visit
  result.get(level).add(root.val);
  // children
  for (Node n : root.children) {
    preorder(n, result, level + 1); // no need to check null
  }  
}

Time: $O(N)$
Space: $O(N)$ to store all nodes.

Iteration

In the foreach statement, if p.children is null, it would crash. However, if a node is in p.children, it can’t be null.

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public List<List<Integer>> levelOrder(Node root) {
  if (root == null) {
    return new ArrayList<>();
  }
  List<List<Integer>> result = new ArrayList<>();
  Queue<Node> queue = new LinkedList<>();
  queue.offer(root);
  int level = 0;

  while (queue.size() > 0) {
    // expand result
    result.add(new ArrayList<>());  
    // queue consists all nodes in the current level
    int levelSize = queue.size();
    for (int i = 0; i < levelSize; ++i) {
      Node p = queue.poll();
      result.get(level).add(p.val);
      // offer children into queue
      for (Node child : p.children) {
        queue.offer(child); // no null check
      }
    }
    ++level;
  }
  return result;
}

Time: $O(N)$
Space: $O(N)$ to store all nodes.