Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
Solution:
/** * 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) { int len = 0; ListNode *p = head; while(p){ len++; p = p->next; } return sortedListToBST(head,0,len-1); } private: TreeNode *sortedListToBST(ListNode*& head,int l,int h){ if(l > h) return NULL; int mid = l + (h-l)/2; TreeNode *lChiled = sortedListToBST(head,l,mid-1); TreeNode *p = new TreeNode(head->val); p->left = lChiled; head = head->next; p->right = sortedListToBST(head,mid+1,h); return p; } };
这种bottom-up 方法很巧妙
Time complexity:O(N)
http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html