解题思路：

不能在发现节点不在范围时直接return null，还需要继续向下遍历，因为它的子节点可能在范围内，也可以重新作为新树的一部分

class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        return recur(root, low, high);
    }

    public TreeNode recur(TreeNode root, int low, int high) {
        if (root == null) return null;
        if (root.val < low) {
            TreeNode right = recur(root.right, low, high);
            return right;
        }
        if (root.val > high) {
            TreeNode left = recur(root.left, low, high);
            return left;
        }
        root.left = recur(root.left, low, high);
        root.right = recur(root.right, low, high);
        return root;
    }
}