Inorder traversal a binary tree

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (root != null) {
            ArrayList<Integer> left = inorderTraversal(root.left);
            ArrayList<Integer> right = inorderTraversal(root.right);
            result.addAll(left);
            result.add(root.val);
            result.addAll(right);
        }
        return result;
    }
    //iterative
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        Stack<treenode> stack = new Stack<treenode>();

        while (root != null || !stack.isEmpty()) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                //when root gets to null, pop the stack and move to the right child
                root = stack.pop();
                result.add(root.val);
                root = root.right;
            }
        }
        return result;
    }
}

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """

        ret, stack = [], []

        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                ret.append(root.val)
                root = root.right
        return ret