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

For example, given a 3-ary tree:

Return its preorder traversal as: [1,3,5,6,2,4].

Note:

Recursive solution is trivial, could you do it iteratively?

M1: recursion

time: O(n), space: O(height)

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> preorder(Node root) {
        List
 
<Integer> res = new ArrayList<>();
        preorder(root, res);
        return res;
    }
    
    public void preorder(Node root, List<Integer> res) {
        if(root == null) {
            return;
        }
        res.add(root.val);
        for(int i = 0; i < root.children.size(); i++) {
            preorder(root.children.get(i), res);
        }
    }
}
 

M2: iteration

由於stack是先進後出，在把children加入stack的時候先reverse，使得稍後訪問stack的時候，children的順序是從左到右的

time: O(n), space: O(n)　　-- worst case

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> preorder(Node root) {
        List<Integer> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        Stack<Node> s = new Stack<>();
        
        s.add(root);
        while(!s.isEmpty()) {
            Node tmp = s.pop();
            res.add(tmp.val);
            Collections.reverse(tmp.children);
            for(Node n : tmp.children) {
                if(n != null) {
                    s.push(n);
                }
            }
        }
        return res;
    }
}