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.
从给定的有序链表生成平衡二叉树。
解题思路：最容易想到的就是利用数组生成二叉树的方法，找到中间节点作为二叉树的root节点，然后分别对左右链表递归调用分别生成左子树和右子树。时间复杂度O(N*lgN)

public class Solution {

    public TreeNode sortedListToBST(ListNode head) {
        return toBST(head,null);
    }
    private TreeNode toBST(ListNode head ,ListNode tail){
        if(head == tail) return null;
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next!=tail&& fast.next.next != tail) {
            fast = fast.next.next;
            slow = slow.next;
        }
        TreeNode root = new TreeNode(slow.val);
        root.left = toBST(head,slow);
        root.right = toBST(slow.next,tail);
        
        return root;
    }

}

 空间为O(1)时间为O(n)：

TreeNode *sortedListToBST(ListNode *head)  
{  
    int len = 0;  
       ListNode * node = head;  
       while (node != NULL)  
       {  
           node = node->next;  
           len++;  
       }  
       return buildTree(head, 0, len-1);  
   }  
     
   TreeNode *buildTree(ListNode *&node, int start, int end)  
   {  
       if (start > end) return NULL;  
       int mid = start + (end - start)/2;  
       TreeNode *left = buildTree(node, start, mid-1);  
       TreeNode *root = new TreeNode(node->val);  
       root->left = left;  
       node = node->next;  
       root->right = buildTree(node, mid+1, end);  
       return root;  
   }  

参考：

http://blog.csdn.net/salutlu/article/details/24502109

http://blog.csdn.net/worldwindjp/article/details/39722643

https://www.nowcoder.com/questionTerminal/86343165c18a4069ab0ab30c32b1afd0