590. N 叉树的后序遍历

给定一个 n 叉树的根节点 root ，返回 其节点值的 后序遍历 。

n 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[5,6,3,2,4,1]

示例 2：

输入：root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出：[2,6,14,11,7,3,12,8,4,13,9,10,5,1]

提示：

• 节点总数在范围 [0, 104] 内
• 0 <= Node.val <= 104
• n 叉树的高度小于或等于 1000

进阶：递归法很简单，你可以使用迭代法完成此题吗?

解法思路：

1、递归（Recursion）

2、迭代（Iterator）效率低

法一：

/*
// Definition for a Node.
class Node {public int val;public List<Node> children;public Node() {}public Node(int _val) {val = _val;}public Node(int _val, List<Node> _children) {val = _val;children = _children;}
};
*/class Solution {public List<Integer> postorder(Node root) {// Recursion// Time: O(n), n 为节点数// Space: O(n)List<Integer> res = new ArrayList<>();helper(root, res);return res;}private void helper(Node root, List<Integer> res) {if (root == null) return;for (Node child : root.children) {helper(child, res);}res.add(root.val);}
}

法二：

/*
// Definition for a Node.
class Node {public int val;public List<Node> children;public Node() {}public Node(int _val) {val = _val;}public Node(int _val, List<Node> _children) {val = _val;children = _children;}
};
*/class Solution {public List<Integer> postorder(Node root) {// Iterator// Time: O(n), n 为节点数// Space: O(n)List<Integer> res = new ArrayList<>();if (root == null) return res;Map<Node, Integer> map = new HashMap<Node, Integer>();Deque<Node> stack = new ArrayDeque<Node>();Node node = root;while (node != null || !stack.isEmpty()) {while (node != null) {stack.addFirst(node);List<Node> children = node.children;if (!children.isEmpty()) {map.put(node, 0);node = children.get(0);} else {node = null;}}node = stack.peek();int idx = map.getOrDefault(node, -1) + 1;List<Node> children = node.children;if (!children.isEmpty() && children.size() > idx) {map.put(node, idx);node = children.get(idx);} else {res.add(node.val);stack.removeFirst();map.remove(node);node = null;}}return res;}
}

优化迭代：

/*
// Definition for a Node.
class Node {public int val;public List<Node> children;public Node() {}public Node(int _val) {val = _val;}public Node(int _val, List<Node> _children) {val = _val;children = _children;}
};
*/class Solution {public List<Integer> postorder(Node root) {// Optimize Iterator// Time: O(n), n 为节点数// Space: O(n)List<Integer> res = new ArrayList<>();if (root == null) return res;Deque<Node> stack = new ArrayDeque<>();Set<Node> visited = new HashSet<Node>();stack.addFirst(root);while (!stack.isEmpty()) {Node node = stack.peek();if (node.children.isEmpty() || visited.contains(node)) {stack.removeFirst();res.add(node.val);continue;}for (int i = node.children.size() - 1; i >= 0; --i) {stack.addFirst(node.children.get(i));}visited.add(node);}return res;}
}