• 最近公共父节点（两个节点，都可能为null，都可能不在树上。）

public static TreeNode getParent(TreeNode root, TreeNode node1,TreeNode node2) {
        if(root == null || node1 == null ||  node2 == null) return null;
        //这里可以换成if(root == node1 || root == node2),我只是为了方便测试才这样写
        if(root.val == node1.val || root.val == node2.val) return root;
        TreeNode left = getParent(root.left,node1,node2);
        TreeNode right = getParent(root.right,node1,node2);
        //如果左右子树都能找到，那么当前节点就是最近的公共祖先节点
        if(left != null && right != null) return root;
        //如果左子树上没有，那么返回右子树的查找结果
        if(left == null) return right;
        //否则返回左子树的查找结果
        else return left;
    }



func reverseList02(_ head:ListNode?) -> ListNode? {
    if head == nil || head?.next == nil {
        return head
    }
    
    let temp = head?.next
    let newHead:ListNode? = reverseList02(temp)
    
    temp?.next = head
    head?.next = nil
    
    return newHead
}