590. N-ary Tree Postorder Traversal

Given 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).

 

Follow up:

Recursive solution is trivial, could you do it iteratively?

 

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

class Solution {
    List<Integer> list = new ArrayList<>();
    public List<Integer> postorder(Node root) {
        if (root == null)
            return list;
        
        for(Node node: root.children)
            postorder(node);
        
        list.add(root.val);
        
        return list;
    }
}

class Solution {
    public List<Integer> postorder(Node root) {
        if(root == null) return Collections.emptyList();
        LinkedList<Integer> list = new LinkedList<>();
        Stack<Node> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            Node n = stack.pop();
            if(n.children != null){
                for(Node node: n.children){
                    stack.push(node);
                }
            }
            list.addFirst(n.val);
        }
        
        return list;
    }
}