Insertion Sort List and Binary Tree Paths

题意如下：
Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

1
/ \
2 3
\
5

All root-to-leaf paths are:

[“1->2->5”, “1->3”]，

返回所有的从叶子节点到根节点的路径，这个用递归法，一层层的遍历即可，遍历过程中非空即存入对应的vector中：

class  Solution {private : vector<string>  res;public :   vector<string> binaryTreePaths(TreeNode* root){      if(root == NULL) {           ;      } else if(root->left == NULL && root->right == NULL) {       string temp = to_string(root->val);       res.push_back(temp);       } else{           if(root->left !=NULL){                 traverse(root->left,to_string(root->val));           }          if(root->right !=NULL){            traverse(root->right,to_string(root->val));          }       }       return  res;    }public :  void  traverse(TreeNode*  node, string s){         if(node == NULL)             return;         if(node->left == NULL && node->right == NULL){           string tmp = to_string(node->val);            res.push_back(s+"->"+tmp);            return ;        } else {             if(node->left !=NULL){             string tmp1 = to_string(node->val);             traverse(node->left,s+"->"+tmp1);             }             if(node->right !=NULL){             string tmp2 = to_string(node->val);             traverse(node->right,s+"->"+tmp2);             }        }     }  };        

齐活了，完成。

再来看看链表的，insertion sort List.很明白，用插入算法来排序

没什么可讲的直接上代码：

class Solution {public :  ListNode*  insertionSortList(ListNode*  head){      if(head == NULL || head->next ==NULL)          return head;          ListNode*   p = new ListNode(-1);          p->next = head;         ListNode*  pre=  head;         ListNode*  cur = pre->next;         while(cur !=NULL){               if(cur->val >=pre->val){                  cur = cur->next;                  pre = pre->next;                }else{                   ListNode*  insertpre = p;                   ListNode*  insertcur = p->next;                while(insertcur->val < cur->val){                   insertpre = insertcur;                   insertcur = insertcur->next;                }                pre->next = cur->next;                cur->next = insertcur;                insertpre ->next = cur;                cur = pre->next;         }         }         head = p->next;         return head;      }  };

齐活了。