leetcode

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

/**
 * 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) {}
 * };
 */
struct ListNode
{
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) :val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
		int n = 0;
		ListNode *p = head;
		while(p != NULL)
		{
			p = p->next;
			n++;
		}
		return ListToBST(head,0,n-1);
    }
private:
	TreeNode *ListToBST(ListNode *&head,int start, int end)
	{
		if(start > end) return NULL;
		int mid = (start + end) / 2;
		TreeNode *left = ListToBST(head,start,mid - 1);
		TreeNode *node = new TreeNode(head->val);
		node->left = left;
		head = head->next;
		node->right = ListToBST(head,mid + 1,end);
		return node;
	}
};


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原文地址:https://www.cnblogs.com/blfshiye/p/4749666.html