算法题收藏

1、Given a BST (Binary search Tree) how will you find median in that?
 
Constraints:
 
* No extra memory.
* Function should be reentrant (No static, global variables allowed.)
* Median for even no of nodes will be the average of 2 middle elements and for odd no of terms will be middle element only.
* Algorithm should be efficient in terms of complexity. 

 

 

思路：

   1）中序遍历搜索二叉树得到有序的双向链表

   2）采用快慢指针来寻找链表的中间元素

 

代码：

typedef struct node{int data;node * lChild;node * rChild;}myNode;// 二叉搜索树node －> 有序的双向链表pre， pre指向链表中最后一个元素void BST2DLL(myNode *node, myNode* &pre){if(NULL == node){return;}BST2DLL(node->lChild, pre);node->lChild = pre;if(NULL != pre){pre->rChild = node;}pre = node;BST2DLL(node->rChild, pre);}// 寻找双向链表pNode的中间元素，其中pNode指向链表中的最后一个元素int FindMid(myNode* pNode){assert(NULL != pNode);myNode* pFast = NULL;myNode* pSlow = NULL;pFast = pNode;pSlow = pNode;while(NULL != pFast && NULL != pSlow){if(NULL == pFast->lChild){return pSlow->data;}else if(NULL == pFast->lChild->lChild){return (pSlow->data + pSlow->lChild->data) >> 1;}else{pFast = pFast->lChild->lChild;pSlow = pSlow->lChild;}}}

 

 

参考：http://blog.csdn.net/hhygcy/archive/2009/10/11/4654305.aspx