LeetCode | Convert Sorted List to Binary Search Tree

2019年3月15日 161次阅读 来源: Allanxl

题目：

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


思路：

类似
http://blog.csdn.net/lanxu_yy/article/details/11898523，不过需要利用链表的方式找到中间节点。

代码：

/**
 * 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:
    ListNode* getMid(ListNode *head, ListNode *tail)
    {
        if(head == tail)
            return NULL;
        else
        {
            ListNode * slow=head;
            ListNode * fast=head;
            while(fast != tail && fast->next != tail)
            {
                slow = slow->next;
                fast = fast->next->next;
            }
            return slow;
        }
    }
    
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return sortedArrayToBST(head, NULL);
    }
    
    TreeNode *sortedArrayToBST(ListNode *head, ListNode *tail)
    {
        
        ListNode * p = getMid(head, tail);
        if(p == NULL)
        {
            return NULL;
        }
        else
        {
            TreeNode * parent = new TreeNode(p->val);
            parent->left = sortedArrayToBST(head, p);
            parent->right = sortedArrayToBST(p->next, tail);
            return parent;
        }
    }
};
    原文作者：Allanxl
    原文地址: https://blog.csdn.net/lanxu_yy/article/details/11900607
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