给出一个所有元素以升序排序的单链表，将它转换成一棵高度平衡的二分查找树
您在真实的面试中是否遇到过这个题？
分析：就是一个简单的递归，只是需要有些链表的操作而已
代码：
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/** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */class Solution { public: /** * @param head: The first node of linked list. * @return: a tree node */ TreeNode *sortedListToBST(ListNode *head) { // write your code here if(head==nullptr) return nullptr; int len = 0; ListNode*temp = head; while(temp){len++;temp = temp->next;}; if(len==1) { return new TreeNode(head->val); } else if(len==2) { TreeNode*root = new TreeNode(head->val); root->right = new TreeNode(head->next->val); return root; } else { len/=2; temp = head; int cnt = 0; while(cnt<len) { temp = temp->next; cnt++; } ListNode*pre = head; while(pre->next!=temp) pre = pre->next; pre->next = nullptr; TreeNode*root = new TreeNode(temp->val); root->left = sortedListToBST(head); root->right = sortedListToBST(temp->next); return root; } } }; |
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原文链接：http://blog.csdn.net/wangyuquanliuli/article/details/47359405
