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Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
把有序链表转化成平衡的BST
/**
* 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;
vector<int> v;
ListNode* p=head;
while(p)
{
v.push_back(p->val);
p=p->next;
}
return BuildTree(v,0,v.size()-1);
}
TreeNode* BuildTree(vector<int>& nums,int first,int end)
{
if(first>end)return NULL;
if(first==end)return new TreeNode(nums[first]);
int mid=(first+end)/2;
TreeNode* root=new TreeNode(nums[mid]);
root->left=BuildTree(nums,first,mid-1);
root->right=BuildTree(nums,mid+1,end);
return root;
}
};leetcode No109. Convert Sorted List to Binary Search Tree
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原文地址:http://blog.csdn.net/u011391629/article/details/52296942