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.

主要想好返回值，每层的返回的东西

思路一： 就是要每次遍历两遍来找中间值。

思路二： 用双指针，第一个指针每次走两步，第二个指针每次走一步，然后当第一个指针走到末尾的时候，第二个就是中间值。

思路三： 一次遍历，先构建左子数，然后构建list上的node往右走。

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if(head==NULL)
            return NULL;
        ListNode *p = head; 
        TreeNode *root = new TreeNode(head->val);

        if(head->next == NULL)
            return root; 
        int length = 0;
        while(p)
        {
            length ++;
            p = p->next; 
        }

        ListNode *fakeHead = new ListNode(0);
        fakeHead->next = head; 
        ListNode * p1 = head; 
        ListNode * pre = fakeHead;
        int mid = 1;
        while(mid != (length+1)/2)
        {
            mid++;
            pre = pre->next; 
            p1 = p1->next; 
        }
        pre->next = NULL;
        root->val = p1->val; 
        ListNode *first = head; 
        ListNode *second = p1->next; 
        p1->next = NULL;
        
        root->right = sortedListToBST(second);
        p1->next = NULL;
        if(length>=3)
            root->left = sortedListToBST(first);
        return root;   
    }
};

思路三：

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        int listLen = 0;
        ListNode *cur = head;
        while(cur) {
            listLen++;
            cur = cur->next;
        }
        return sortedListToBST(head, 0, listLen-1);
    }
    
    TreeNode *sortedListToBST(ListNode *&head, int start, int end) {
        if(start>end) return NULL;
        int mid = start + (end-start)/2;
        TreeNode *leftChild = sortedListToBST(head, start, mid-1);
        TreeNode *root = new TreeNode(head->val);
        head = head->next;
        TreeNode *rightChild = sortedListToBST(head, mid+1, end);
        root->left = leftChild;
        root->right = rightChild;
        return root;
    }
};