该题将有序链表转为二叉查找树,通过类似有序数组的方法,每次找到中间值作为根,然后递归求解分开的两个子链,代码如下
void myf(TreeNode * &T,ListNode *b,ListNode *e){ if(b->next==e){T=new TreeNode(b->val);return;} ListNode *mid=b; ListNode *tail=b->next; while(tail!=e&&tail->next!=e){tail=tail->next->next;mid=mid->next;} T =new TreeNode(mid->val); if(mid!=b) myf(T->left,b,mid); if(mid!=e) myf(T->right,mid->next,e); } class Solution { public: TreeNode *sortedListToBST(ListNode *head) { TreeNode *root=NULL; if(head==NULL)return root; ListNode *tail=head; while(tail)tail=tail->next; myf(root,head,tail); return root; } };