Sword finger offer merges two sorted lists

describe
Enter two incremental linked lists , The length of a single linked list is n, Merge these two linked lists and make the nodes in the new linked list still be sorted incrementally .

Such as the input {1,3,5},{2,4,6} when , The combined linked list is {1,2,3,4,5,6}, So the corresponding output is {1,2,3,4,5,6}, The conversion process is shown in the figure below ：

Their thinking ：
A special case ： If a linked list is empty , Return to another linked list
If pHead1 Node value ratio pHead2 Small , The next node should be pHead1, should return pHead1, stay return Before , Appoint pHead1 The next node of should be pHead1.next and pHead2 The merged head node of the two linked lists
If pHead1 Node value ratio pHead2 Big , The next node should be pHead2, should return pHead2, stay return Before , Appoint pHead2 The next node of should be pHead1 and pHead2.next The merged head node of the two linked lists
Python Code implementation

class Solution:
    #  Return to merged list 
    def Merge(self, pHead1, pHead2):
        #  The end condition of recursion is that there is an end , Return no None That's good 
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        #  Give Way phead1 Point to the small , Exchange if necessary 
        if pHead1.val > pHead2.val:
            pHead1,pHead2 = pHead2,pHead1
        #  recursive procedure 
        pHead1.next = self.Merge(pHead1.next,pHead2)
        return pHead1