leetcode-590-easy

N-ary Tree postorder Traversal

Given the root of an n-ary tree, return the postorder 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)

Example 1:

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

The number of nodes in the tree is in the range [0, 104].
0 <= Node.val <= 104
The height of the n-ary tree is less than or equal to 1000.
Follow up: Recursive solution is trivial, could you do it iteratively?

思路一：后续遍历，用递归实现

    private final List<Integer> postOrderList = new ArrayList<>();
    public List<Integer> postorder(Node root) {
        if (root == null) {
            return new ArrayList<>();
        }

        for (Node child : root.children) {
            postorder(child);
        }
        postOrderList.add(root.val);

        return postOrderList;
    }