如何使用递归返回从树根到目标的树节点的ArrayList？

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    
    public TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

public class TreeTraversal {
    public static ArrayList<Integer> findPath(TreeNode root, TreeNode target) {
        ArrayList<Integer> path = new ArrayList<>();
        if (root == null) {
            return path;
        }
        
        if (root == target) {
            path.add(root.val);
            return path;
        }
        
        ArrayList<Integer> leftPath = findPath(root.left, target);
        if (!leftPath.isEmpty()) {
            path.addAll(leftPath);
            path.add(root.val);
            return path;
        }
        
        ArrayList<Integer> rightPath = findPath(root.right, target);
        if (!rightPath.isEmpty()) {
            path.addAll(rightPath);
            path.add(root.val);
            return path;
        }
        
        return path;
    }
    
    public static void main(String[] args) {
        // 创建一个示例树
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);
        root.right.left = new TreeNode(6);
        root.right.right = new TreeNode(7);
        
        // 查找从根节点到目标节点5的路径
        TreeNode target = root.left.right;
        ArrayList<Integer> path = findPath(root, target);
        
        // 输出路径
        for (int i = path.size() - 1; i >= 0; i--) {
            System.out.print(path.get(i) + " ");
        }
    }
}