Description： Given an n-ary tree, return the postorder traversal of its nodes’ values.

For example, given a 3-ary tree:

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

Note: Recursive solution is trivial, could you do it iteratively?

題意：給定一顆N叉樹，要求返回後序遍歷的節點值；

解法一（遞迴）：我們可以很容易的想到用遞迴來解決樹的遍歷問題；對於每一個節點，我們依次訪問他們的每一個子節點，對於每一個子節點，再遞迴的遍歷自己的子節點；

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> postorder(Node root) {
        LinkedList<Integer> result = new LinkedList<>();
        if (root == null) return result;
        for (int i = 0; i < root.children.size(); i++) {
            postorder(root.children.get(i), result);
        }
        result.addLast(root.val);
        return result;
    }
    
    private void postorder(Node root, LinkedList<Integer> result) {
        if (root.children.size() == 0) {
            result.addLast(root.val);
            return;
        }
        for (int i = 0; i < root.children.size(); i++) {
            postorder(root.children.get(i), result);
        }
        result.add(root.val);
    }
}
 

解法二（非遞迴）：如果要利用非遞迴來解決這個問題的話，我們需要藉助一個棧來儲存子節點，對於每一個節點，我們將其所有的子節點都入棧；每一次，我們都從棧中取出一個節點，將其插入到list的首部（因為棧是後進先出的，所有第一個子節點是先入棧最後出棧的）；

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> postorder(Node root) {
        LinkedList<Integer> result = new LinkedList<>();
        Stack<Node> treeNode = new Stack<>();
        if (root == null) return result;
        treeNode.push(root);
        while (!treeNode.isEmpty()) {
            Node temp = treeNode.pop();
            result.addFirst(temp.val);
            for (Node node : temp.children) {
                treeNode.push(node);
            }
        }
        return result;
    }
}