11
08/2015
[LeetCode] Convert Sorted List to Binary Search Tree
Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
解题思路：
题意为构造有序链表的二分查找树。找到中间节点的办法用双指针法。注意我们还需要保存中间节点的前一个节点，便于一个链表分成两个链表。注意滴38行，是将链表恢复成原来的模样。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* sortedListToBST(ListNode* head) {
if(head == NULL){
return NULL;
}
ListNode* myHead = new ListNode(0);
myHead->next = head;
ListNode* pre = myHead;
ListNode* one = head, *two = head;
while(two->next!=NULL && two->next->next!=NULL){
pre = pre->next;
one = one->next;
two = two->next->next;
}
two = one->next;
pre->next = NULL;
TreeNode* root = new TreeNode(one->val);
root->left = sortedListToBST(myHead->next);
root->right = sortedListToBST(two);
pre->next = one; //恢复链表结构
//delete one; //这里为啥不能删除？
delete myHead;
return root;
}
};
0 条评论