Convert Sorted List to Binary Search Tree

一、问题描述

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

二、思路

数组的关键在于下标，链表的关键在于首位节点。我们递归的退出条件是头尾节点相等：

        if(head == tail)
            return NULL;

即链表为空，另一个退出条件是链表只有一个节点：

        if(head -> next == tail){
            TreeNode *root = new TreeNode(head->val);
    		return root;
        }

最关键的部分在于找到中间节点，和数组转变为二叉树思路类似。Convert Sorted Array to Binary Search Tree

为了找到中间节点，我们用了快慢指针，mid和temp。

三、代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        return createTree(head,NULL);
    }
    TreeNode* createTree(ListNode* head, ListNode* tail){
        if(head == tail)
            return NULL;
        if(head -> next == tail){
            TreeNode *root = new TreeNode(head->val);
    		return root;
        }
        ListNode *mid = head, *temp = head;
        while(temp != tail && temp->next != tail){
    		mid = mid->next;
    		temp = temp->next->next;
    	}
    	TreeNode *root = new TreeNode(mid->val);
    	root->left = createTree(head, mid);
    	root->right = createTree(mid->next, tail);
    	return root;
    }
};