109. Convert Sorted List to Binary Search Tree
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109. Convert Sorted List to Binary Search Tree
题目
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
解析
- 主要考察链表求中间节点
- 统一输入输出接口,需要将前后两段指针分开,也可以再加入一个尾指针参数
- 这个题是这几天做的最顺利的,一次AC掉
/**
* 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) {
ListNode* slow = head;
ListNode* fast = head;
ListNode* pre = NULL;
if (!head)
{
return NULL;
}
if (!head->next)
{
TreeNode* temp = new TreeNode(head->val);
return temp;
}
while (fast&&fast->next)
{
pre = slow;
slow = slow->next;
fast = fast->next->next;
}
TreeNode* root = new TreeNode(slow->val);
pre->next = NULL;
slow = slow->next;
root->left = sortedListToBST(head);
root->right = sortedListToBST(slow);
return root;
}
};
题目来源
C/C++基本语法学习
STL
C++ primer