106. 排序列表转换为二分查找树
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
/**
* 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;
}
}
};