590. N叉树的后序遍历

给定一个 N 叉树，返回其节点值的后序遍历。

例如，给定一个 3叉树 :

narytreeexample.png

返回其后序遍历: [5,6,3,2,4,1].

说明: 递归法很简单，你可以使用迭代法完成此题吗?
题解：此题与589同源，589是前序，故先入根节点，590后序最后入根节点即可。

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

2.迭代：

public List<Integer> postorder2(Node root) {
    List<Integer> res = new ArrayList<>();
    if (root == null) return res;
    //前指针
    Node pre = null;
    Stack<Node> stack = new Stack<>();
    stack.push(root);
    while (!stack.isEmpty()) {
        Node cur = stack.peek();
        if ((cur.children.size() == 0) || (pre != null && cur.children.contains(pre))) {
            //加入结果集
            res.add(cur.val);
            stack.pop();
            //更新pre指针
            pre = cur;
        } else {
            //继续压栈，注意压栈是从右往左
            List<Node> nodeList = cur.children;
            for (int i = nodeList.size() - 1; i >= 0; i--) {
                stack.push(nodeList.get(i));
            }
        }
    }
    return res;
}