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Similar to the sorted array. But we have to remember, the position of node not depends on the index. So we need do two more things:
1. Ensure the pointer is moving (change value) in function call. So we need pass the pointer‘s pointer to the function.
2. Get left branch first, so the pointer can move to the root node now.
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 /** 10 * Definition for binary tree 11 * struct TreeNode { 12 * int val; 13 * TreeNode *left; 14 * TreeNode *right; 15 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 16 * }; 17 */ 18 class Solution { 19 public: 20 TreeNode *getTree(ListNode **head, int start, int end) { 21 if (start > end) return NULL; 22 int mid = (start + end)/2; 23 TreeNode *lnode = getTree(head, start, mid-1); 24 TreeNode *root = new TreeNode((*head)->val); 25 *head = (*head)->next; 26 root->left = lnode; 27 root->right = getTree(head, mid+1, end); 28 return root; 29 } 30 TreeNode *sortedListToBST(ListNode *head) { 31 if (!head) return NULL; 32 ListNode *runner = head; 33 int index = 0; 34 while (runner->next) { 35 runner = runner->next; 36 index++; 37 } 38 return getTree(&head, 0, index); 39 } 40 };
LeetCode – Refresh – Convert Sorted List to Binary Search Tree
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原文地址：http://www.cnblogs.com/shuashuashua/p/4349281.html
