589. N-ary Tree Preorder Traversal.md

589. N-ary Tree Preorder Traversal

Given an n-ary tree, return the preorder traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Follow up:

Recursive solution is trivial, could you do it iteratively?

Example 1:

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

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]

Constraints:

• The height of the n-ary tree is less than or equal to 1000
• The total number of nodes is between [0, 10^4]

Solution

• 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;
      }
  };
  */
  

  • mine Runtime: 2 ms, faster than 48.85%, Memory Usage: 40.4 MB, less than 100.00% of Java online submissions

  // O(N)time O(N)space, N is the node count.
  class Solution {
      public List<Integer> preorder(Node root) {
          List<Integer> res = new ArrayList<>();
          if(root == null){
              return res;
          }
          res.add(root.val);
          if(root.children != null  && root.children.size() > 0){
              for(Node n : root.children){
                  res.addAll(preorder(n));
              }
          }
          return res;
      }
  }