Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
/**
* 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)
{
if(head == NULL)
return NULL;
vector<int> res;
ListNode* pre = head;
while(pre!=NULL)
{
res.push_back(pre->val);
pre = pre->next;
}
int n=res.size();
return sortedArrayToBST(res,0,n-1);
}
TreeNode *sortedArrayToBST(vector<int> &num, int low, int high)
{
if(low>high)
return NULL;
if(low == high)
{
return new TreeNode(num[low]);
}
int mid = (low+high)/2;
TreeNode* root = new TreeNode(num[mid]);
root->left = sortedArrayToBST(num,low,mid-1);
root->right = sortedArrayToBST(num,mid+1,high);
return root;
}
};LeetCode--Convert Sorted List to Binary Search Tree
原文地址:http://blog.csdn.net/shaya118/article/details/42710239