590. N-ary Tree Postorder Traversal

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LeetCode

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Problem

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 $[0, 10^4]$.
• $0 <= Node.val <= 10^4$
• 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?

Code

/*
// 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;
    }
};
*/

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

/*
// 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;
    }
};
*/

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

        Collections.reverse(list);

        return list;
    }
}