Description： 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?

題意：返回一顆N叉樹的前序遍歷節點值；

解法一（遞迴）：將根節點的值插入到list中，對根節點，我們依次遍歷他的所有子節點，對每一個子節點遞迴呼叫自身；

Java

/*
// 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 {
    List<Integer> result = new ArrayList<>();
    public List<Integer> preorder(Node root) {
        if (root == null) return result;
        result.add(root.val);
        for (Node node : root.children) {
            preorder(node);
        }
        return result;
    }
}
 

解法二（非遞迴）：要想使用非遞迴實現前序遍歷，我們需要利用一個棧來儲存子節點；對每一個節點，我們從他的最後節點依次到最左節點存入到棧中；每一次，我們取出一個棧中元素，將其值插入到list中，再將其所有子節點從右往左依次儲存到棧中；直到最後的棧為空；

Java

/*
// 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> result = new ArrayList<>();
        Stack<Node> treeNode = new Stack<>();
        if (root == null) return result;
        treeNode.push(root);
        while (!treeNode.isEmpty()) {
            Node temp = treeNode.pop();
            result.add(temp.val);
            for (int i = temp.children.size() - 1; i >= 0; i--) {
                treeNode.push(temp.children.get(i));
            }
        }
        return result;
    }
}