106. 排序列表转换为二分查找树

Posted 2020-10-17 Carlos

106. 排序列表转换为二分查找树 

 

•  描述
•  笔记
•  数据
•  评测

给出一个所有元素以升序排序的单链表，将它转换成一棵高度平衡的二分查找树

您在真实的面试中是否遇到过这个题？ 

Yes

样例

               2
1->2->3  =>   /              1   3

标签 

链表 递归

 

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */


class Solution {
public:
       /**
     * @param head: The first node of linked list.
     * @return: a tree node
     */
    TreeNode *sortedListToBST(ListNode *head) {
        // write your code here
     	//用递归 
    	TreeNode* root = nullptr;
    	if (head != nullptr)
    	{
    		ListNode *left = nullptr, *right = nullptr;
    		root = new TreeNode(dichotomyList(head, left, right));
    		root->left = sortedListToBST(left);
    		root->right = sortedListToBST(right);
    	}
    }

    int dichotomyList(ListNode *head, ListNode *&left, ListNode *&right) {
       	if (head->next != nullptr)
    	{
    		ListNode *fast = head, *slow = head, *temp = head;
    		while (fast != nullptr && fast->next != nullptr)
    		{
    			temp = slow;
    			slow = slow->next;
    			fast = fast->next->next;
    		}
    		temp->next = nullptr;
    		left = head;
    		right = slow->next;
    		return slow->val;
    	}
    	else
    	{
    		left = nullptr;
    		right = nullptr;
    		return head->val;
    	}
    }
};