[leetcode] 589. N-ary Tree Preorder Traversal

题目：

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?

代码：

/*
// 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) {
        if(root == null) return new ArrayList<Integer>();
        List<Integer> list = new ArrayList<Integer>();
        list.add(root.val);
        Iterator<Node> iter = root.children.iterator();
        while(iter.hasNext()){
            list.addAll(preorder(iter.next()));
        }
        return list;
    }
}

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

    Node() {}

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
public:
    vector<int> preorder(Node* root) {
        if(root == NULL) return vector<int>();
        stack<Node*> s;
        vector<int> res;
        s.push(root);
        while(!s.empty()){
            Node* temp = s.top();
            s.pop();
            res.push_back(temp -> val);
            for(int i = temp -> children.size()-1; i >= 0; i--){
                s.push(temp -> children[i]);
            }
        }
        return res;
    }
};