LintCode-排序列表轉換為二分查找樹分析及實例。本站提示廣大學習愛好者:(LintCode-排序列表轉換為二分查找樹分析及實例)文章只能為提供參考,不一定能成為您想要的結果。以下是LintCode-排序列表轉換為二分查找樹分析及實例正文
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找樹
您在真實的面試中是否遇到過這個題?
分析:就是一個簡單的遞歸,只是需要有些鏈表的操作而已
代碼:
/** * 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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