LeetCode-N-ary Tree Postorder Traversal

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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?

题意：给定一颗N叉树，要求返回后序遍历的节点值；

解法一（递归）：我们可以很容易的想到用递归来解决树的遍历问题；对于每一个节点，我们依次访问他们的每一个子节点，对于每一个子节点，再递归的遍历自己的子节点；

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（非递归）：如果要利用非递归来解决这个问题的话，我们需要借助一个栈来存储子节点，对于每一个节点，我们将其所有的子节点都入栈；每一次，我们都从栈中取出一个节点，将其插入到list的首部（因为栈是后进先出的，所有第一个子节点是先入栈最后出栈的）；

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;
    }
}