145 二叉树的后序遍历

输入: [1,null,2,3]  
   1
    
     2
    /
   3 

输出: [3,2,1]

递归实现：
public List<Integer> postorderTraversal(TreeNode root) {

        List<Integer> res = new ArrayList<>();
        postOrder2(root, res);
        retu-rn res;
    }

    public void postOrder2(TreeNode root, List<Integer> res) {
        if (root.left != null) {
            postOrder2(root.left, res);
        }
        if (root.right != null) {
            postOrder2(root.right, res);
        }
        res.add(root.val);
    }

非递归实现：
 public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Stack<Integer> res = new Stack<>();
        Stack<TreeNode> stack = new Stack<>();
        if (root == null) {
            return res;
        }
        TreeNode node = root;
        while (node != null || !stack.empty()) {
            if (node != null) {
                res.push(node.val);
                stack.push(node);
                node = node.right;
            } else {
                node = stack.pop();
                node = node.left;
            }
        }
        while(!res.empty()){
            result.add(res.pop());
        }
        return result;
    }