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.
分析: 根据一个有序链表,得到一个平衡二叉搜索树,主要是根据快慢指针得到链表的中点,然后将链表分成两半
/**
* 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:
ListNode* findMiddle(ListNode* start){
ListNode *low = start, *fast = start;
ListNode* prelow=nullptr;
while(fast!=nullptr){
fast = fast->next;
if(fast){
fast = fast->next;
prelow = low;
low = low->next;
} }
if(prelow)
prelow->next =nullptr;//break the list
return low;
} TreeNode* sortedListToBST(ListNode* head) {
if(head==nullptr)
return nullptr;
if(head->next ==nullptr)
return new TreeNode(head->val);
ListNode* mid = findMiddle(head);
TreeNode* root = new TreeNode(mid->val);
root->left = sortedListToBST(head);
root->right = sortedListToBST(mid->next);
return root;
}
};
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