lintcode 106. 排序列表轉換為二分查詢樹
阿新 • • 發佈:2018-11-10
題目:給出一個所有元素以升序排序的單鏈表,將它轉換成一棵高度平衡的二分查詢樹
思路:每次取一箇中間元素,遞迴。
class Solution {
public:/*
* @param head: The first node of linked list.
* @return: a tree node
*/
TreeNode * func(int left,int right,vector<int> arr){
if(right<left) return NULL;//? <=
int mid=(left+right)/2;
TreeNode *tree = new TreeNode(arr[mid]);
tree->left=func(left,mid-1,arr);
tree->right=func(mid+1,right,arr);
return tree;
}
TreeNode * sortedListToBST(ListNode * head) {
if(head==NULL) return NULL;
vector<int> arr;
ListNode *p=head;
while(p!=NULL){
arr.push_back(p->val);
p=p->next;
}
return func(0,arr.size()-1,arr);
// write your code here
}
};
ps:樹那裡都要忘得差不多了 每天一道lintcode 撿起來