力扣算法题—145BinartTreePostorderTraversal

Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

Solution：

　　使用两个栈，来实现；

　　使用一个栈来实现

　　

 class Solution {
 public:
     ListNode* sortList(ListNode* head) {
         if (head == nullptr || head->next == nullptr)return head;
         ListNode *slow = head, *fast = head, *pre = head;
         while (fast != nullptr && fast->next != nullptr)
         {
             pre = slow;
             slow = slow->next;
             fast = fast->next->next;
         }
         pre->next = nullptr;//将链表分为两段
         return merge(sortList(head), sortList(slow));
     }
     ListNode* merge0(ListNode* l1, ListNode* l2)
     {
         if (l1 == nullptr || l2 == nullptr)return l1 == nullptr ? l2 : l1;
         if (l1->val < l2->val)
         {
             l1->next = merge(l1->next, l2);
             return l1;
         }
         else
         {
             l2->next = merge(l1, l2->next);
             return l2;
         }
     }
     ListNode* merge(ListNode* l1, ListNode* l2)
     {
         ListNode* ptr = new ListNode(-);
         ListNode* p = ptr;
         while (l1 != nullptr && l2 != nullptr)
         {
             if (l1->val < l2->val)
             {
                 p->next = l1;
                 l1 = l1->next;
             }
             else
             {
                 p->next = l2;
                 l2 = l2->next;
             }
             p = p->next;
         }
         if (l1 == nullptr)p->next = l2;
         if (l2 == nullptr)p->next = l1;
         return ptr->next;
     }
 };