Binary Tree Postorder Traversal leetcode

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

postorder： 左－右－中

public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (root == null){
            return result;
        }
        ArrayList<Integer> left = postorderTraversal(root.left);
        ArrayList<Integer> right = postorderTraversal(root.right);
        result.addAll(left);
        result.addAll(right);
        result.add(root.val);
        return result;
        
    }
}

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        helper(res, root);
        return res;
    }
    public void helper(List<Integer> res, TreeNode root) {
        if (root == null) {
            return;
        }
        helper(res, root.left);
        helper(res, root.right);
        res.add(root.val);
    }
}
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        TreeNode pre = null;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while (root != null || !stack.isEmpty()) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                TreeNode peak = stack.peek();
                if (peak.right != null && pre != peak.right) {///如果当前栈顶元素的右结点存在并且还没访问过（也就是右结点不等于上一个访问结点）就访问右结点 /
                    root = peak.right;
                } else {////如果栈顶元素右结点是空或者已经访问过，那么说明栈顶元素的左右子树都访问完毕 需要把栈顶元素加入结果并且回溯上一层
                    stack.pop();
                    res.add(peak.val);
                    pre = peak;
                }
                
            }
        }
        return res;
    }
}