5 Convert Sorted List to Binary Search Tree_Leetcode

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

這題和sort list等題都比較類似，須要先用快慢指針的方法找到鏈表的中點，而後用recursive的方式構建左子樹和右子樹（用到的思想是Divide&Conquer)，而後再構建好這個節點。

編程時一點要注意：

（1）dummy節點的使用能夠幫助找到中點的prev節點

可是dummy節點只有當用時再構造，建議不要提早建好當參數傳進來，會很是不清晰。用完後最後記得delete，不然會memory leak. 也不能delete得過早，由於prev可能等於dummy.

（2）邊界狀況

當middle節點爲head時，應注意，左子樹爲NULL，不要遞歸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;
    }
};