5 Convert Sorted List to Binary Search Tree_Leetcode

这题和sort list等题都比较相似，需要先用快慢指针的方法找到链表的中点，然后用recursive的方式构建左子树和右子树（用到的思想是Divide&Conquer)，然后再构建好这个节点。

编程时一点要注意：

（1）dummy节点的使用可以帮助找到中点的prev节点

但是dummy节点只有当用时再构造，建议不要提前建好当参数传进来，会非常不清晰。用完后最后记得delete，否则会memory leak. 也不能delete得过早，因为prev可能等于dummy.

（2）边界情况

当middle节点为head时，应注意，左子树为NULL，不要递归build。

Code:

/**
* 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 *sortedListToBST(ListNode *head) {
if(!head) return NULL;
TreeNode* treeHead = buildTree(head);
return treeHead;
}

TreeNode* buildTree(ListNode *head)
{
if(!head) return NULL;
if(!head->next) return new TreeNode(head->val);

ListNode* dummy = new ListNode(0);
dummy->next = head;
ListNode* prev = dummy, *slow = dummy->next, *fast = dummy->next->next;
while(fast && fast->next)
{
slow = slow->next;
prev = prev->next;
fast = fast->next->next;
}

TreeNode* cur = new TreeNode(slow->val);

TreeNode* left, *right;  // second error, cannot initialize left in the if-else
right = buildTree(slow->next);
if(prev != dummy)      // first error
{
prev->next = NULL;
left = buildTree(head);
}
else left = NULL;

cur->left = left;
cur->right = right;

delete dummy;  // first error, can not delete dummy too early if prev equals dummy

return cur;
}
};