LeetCode 109 Convert Sorted List to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5


用c++的指针写，还是比较烦的‘
c++

class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        
         if(head==NULL) return NULL;
         int l = 0 ;
         int r = 0 ;
         ListNode* term = head;
         ListNode* left2 = NULL;
         ListNode* right2 = NULL;
        ListNode* left;
        ListNode* right;
        
         while(term!=NULL)
         {
             r++;
             term = term->next;
         }
        
         int mid =(l+r)/2;
         term = head;
         int value;
         l=0;
         while(term!=NULL)
         {
            if(l<mid)
            {
                ListNode* temp = new ListNode(term->val);
                if(left2 == NULL) {left2 = temp;left = left2;}
                else{
               
                left->next = temp;
                    left = left->next;
                }
                
            }
            else if(l>mid)
            {
                ListNode* temp = new ListNode(term->val);
                  if(right2 == NULL) {right2 = temp;right=right2;}
                else{
                  right->next = temp;
                    right = right->next;
                }
            }
            else if(l==mid)
                value = term->val;
            term = term->next;
             l++;
         }
       
        
         TreeNode* tree = new TreeNode(value);
         tree->left = sortedListToBST(left2);
         tree->right = sortedListToBST(right2);
        
        return tree;
        
    }
};