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(-1);
        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;
    }
};