【LeetCode】109&

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

Solution 1:  recursion 

runtime: 28ms.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if(!head)return NULL;
        ListNode *temp=head;
        int length=0;
        while(temp){
            length++;
            temp=temp->next;
        }
        return sortedListToBST(head, 0, length-1);
    }
    TreeNode* sortedListToBST(ListNode *& head, int begin, int end){
        if(begin>end)return NULL;
        int mid=begin+(end-begin)/2;
        TreeNode *left=sortedListToBST(head,begin,mid-1);
        TreeNode *root=new TreeNode(head->val);
        head=head->next;
        TreeNode *right=sortedListToBST(head,mid+1,end);
        root->left=left;
        root->right=right;
        return root;
    }

TreeNode* sortedArrayToBST(vector<int>& nums) {

int n=nums.size();
return sortedArrayToBST(nums, 0, n-1);

}

TreeNode* sortedArrayToBST(vector<int>& nums, int start,int end) {

if (start > end) return NULL;
int mid = start + (end - start) / 2;

TreeNode *node = new TreeNode(nums[mid]);
node->left = sortedArrayToBST(nums, start, mid-1);
node->right = sortedArrayToBST(nums, mid+1, end);
return node;

}

};

Solution 2:https://leetcode.com/discuss/29826/clean-c-solution-recursion-o-nlogn-with-comment