Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode *creatTree(ListNode *&head, int left, int right) { if(left>right)return NULL; int mid=(left+right)/2; TreeNode *treeLeft=creatTree(head,left,mid-1); TreeNode *root=new TreeNode(head->val); root->left=treeLeft; head=head->next; root->right=creatTree(head,mid+1,right); } TreeNode *sortedListToBST(ListNode *head) { TreeNode *root; int len=0; ListNode *tmp=head; while(tmp) { len++; tmp=tmp->next; } root=creatTree(head, 0, len-1); } /* void creatTree(TreeNode *&root,vector<int> &num,int left,int right) { if(left>right)return; int mid=(left+right)/2; root=new TreeNode(num[mid]); creatTree(root->left,num,left,mid-1); creatTree(root->right,num,mid+1,right); } TreeNode *sortedArrayToBST(vector<int> &num) { TreeNode *root=NULL; creatTree(root,num,0,num.size()-1); return root; } TreeNode *sortedListToBST(ListNode *head) { TreeNode *root=NULL; if(head==NULL)return root; vector<int> num; while(head) { num.push_back(head->val); head=head->next; } return sortedArrayToBST(num); } */ };