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？
//leetcode submit region begin(Prohibit modification and deletion)
/**
* 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) {
List<Integer> result = new ArrayList<>();
doPostorder(result, root);
return result;
}
private void doPostorder(List<Integer> result, TreeNode node) {
if (node == null) {
return;
}
doPostorder(result, node.left);
doPostorder(result, node.right);
result.add(node.val);
}
}
//leetcode submit region end(Prohibit modification and deletion)
