题目描述:Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
思路分析:由于给出的单链表是已经排序好的,本题我们只需要找到中间节点,将其标记为树的根,然后分别递归调用左子树和右子树。
代码实现如下:
/**
* 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)
{
//判空
if(head==NULL)
return NULL;
//若节点只有一个
if(head->next==NULL)
return new TreeNode(head->val);
//用快慢指针法求中间节点
ListNode* fast=head;
ListNode* slow=head;
ListNode* cur=head;
while(fast!=NULL && fast->next!=NULL)
{
cur=slow;
slow=slow->next;
fast=fast->next->next;
}
//构造树
TreeNode* root = new TreeNode(slow->val);
cur->next=NULL;
//递归遍历左子树和右子树
root->left=sortedListToBST(head);
root->right=sortedListToBST(slow->next);
return root;
}
};