145. 二叉树的后序遍历

给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。

示例 1：

输入：root = [1,null,2,3]
输出：[3,2,1]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

提示：

• 树中节点的数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

进阶：递归算法很简单，你可以通过迭代算法完成吗？

解法思路：

1、递归

2、迭代

 法一：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<Integer> postorderTraversal(TreeNode root) {// Recursion// Time: O(n)// Space: O(n)List<Integer> res = new ArrayList<>();postorder(root, res);return res;}private void postorder(TreeNode root, List<Integer> res) {if (root == null) return;postorder(root.left, res);postorder(root.right, res);res.add(root.val);}
}

法二：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public List<Integer> postorderTraversal(TreeNode root) {// Iterator// Time: O(n)// Space: O(n)List<Integer> res = new ArrayList<>();if (root == null) return res;Deque<TreeNode> stack = new ArrayDeque<>();TreeNode prev = null;while (root != null || !stack.isEmpty()) {while (root != null) {stack.addLast(root);root = root.left;}root = stack.removeLast();if (root.right == null || root.right == prev) {res.add(root.val);prev = root;root = null;} else {stack.addLast(root);root = root.right;}}return res;}
}