Reorder List and Sum of Left Leaves

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes’ values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

这个题目有点类似于综合题的感觉。对比给出实例。可以看出是将后半部的链表翻转后再与前半部合并。
算法分析：
1. 首先要分解链表，将链表一分为二。
2. 将后半部分链表倒序。
3. 合并两个链表。

class  Solution {public:   void reoderList(ListNode* head){       if(head == NULL)          return ;       ListNode*   p1 = head;       ListNode*    p2 = splitList(head);       p2 = reverseList(p2);       mergelist(p1,p2);   }  ListNode*  splitList(ListNode* head){      ListNode*   slow = new ListNode(0);      slow->next = head;      ListNode*   fast = slow;      while(fast->next && fast->next->next){              slow = slow->next;              fast = fast->next->next;      }     if(fast->next){        slow = slow->next;        fast = fast->next;        }        ListNode*  tmp = slow->next;        slow->next =NULL;         return tmp;   }  ListNode*  reverseList(ListNode*  head){      if(head == NULL)         return head;       ListNode*  p=head;       p = head->next;       head->next = NULL;       while(p){            ListNode*   tmp = p->next;            p->next = head;            head = p;            p = tmp;       }       return head;  }  void  mergelist(ListNode* p1,ListNode*  p2){      while(p2){         ListNode*   tmp = p2;         p2 = p2->next;         p1->next =p2;         tmp->next = p1->next;         p1 = p1->next->next;      }  }};

再来看看二叉树：Sum of Left Leaves
计算所有左子树之和。
这个很容易理解，进行递归求解：

int  sumOfLeftLeaves(ListNode* root){  if(root == NULL)  return 0;if(root->left =NULL && root->left->left =NULL           && root->left->right == NULL){           return root->val + sumOfLeftLeaves(root->right);           }return sumOfLeftLeaves(root->left) +    sumOfLeftLeaves(root->right);}

好了，今天的解析到此为止了。