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. 

中文不需要贅述了，就是二叉搜尋樹如何高效地找到中位數，不希望申請記憶體，不要static/global的變數來實現。

第一反應就是中序遍歷就是排序了，但是如果要計數的話，我們需要一個額外的變數，這樣恐怕需要或是靜態或者全域性變量了，故不可以用。但是我們其他很多次都碰到了那樣類似的問題，如何把一個二叉樹原地轉化成雙向連結串列，那時候還有點覺得題目沒意思，沒啥用處。但是在這裡，作用得以體現。假如有形如

                                            10
                                          /    /
                                        6       14
                                      /  /     /　 /
                                   　4     8  12 　  16

的二叉樹轉化成了4=6=8=10=12=14=16這樣一個雙鏈表，那後面的問題就變成了如何在4=6=8=10=12=14=16裡面找到中間節點了，這對於我們習慣了2個快慢指標追趕的小朋友來說不算問題了。

下面就是寫的一個實現：

#include <iostream> typedef struct bstNode { int data; bstNode* pLeft; bstNode* pRight; bstNode() { data = 0; pLeft = NULL; pRight = NULL; } }bstNode; void bst2dll (bstNode* pNode, bstNode*& pTail ) { // in-order traverse if (pNode == NULL) return ; if (pNode->pLeft) bst2dll(pNode->pLeft,pTail); bstNode* pCurrent = pNode; pCurrent->pLeft = pTail; if (pTail) pTail->pRight = pCurrent; pTail = pCurrent; if(pNode->pRight) bst2dll(pNode->pRight, pTail); } // parameter is the original root // return the new double linked list head int medianInBST ( bstNode* pRoot ) { bstNode* pTail = NULL; bst2dll(pRoot,pTail); // dummy handling here if (pTail == NULL) return -1; bstNode* pFast = pTail; bstNode* pSlow = pTail; while (pFast&&pSlow) { if (pFast->pLeft==NULL) return pSlow->data; else if (pFast->pLeft != NULL && pFast->pLeft->pLeft == NULL) return (pSlow->data + pSlow->pLeft->data)>>1; else { pFast = pFast->pLeft; pFast = pFast->pLeft; pSlow = pSlow->pLeft; } } } // test case /* 10 / / 6 14 / / / / 4 8 12 16 */ bstNode* buildupTree() { // level 1 bstNode* pRoot = new bstNode; pRoot->data = 10; //level2 bstNode* pNewL = new bstNode; pNewL->data = 6; bstNode* pNewR = new bstNode; pNewR->data = 14; //level3 bstNode* pNewLL = new bstNode; pNewLL->data = 4; bstNode* pNewLR = new bstNode; pNewLR->data = 8; bstNode* pNewRL = new bstNode; pNewRL->data = 12; bstNode* pNewRR = new bstNode; pNewRR->data = 16; pRoot->pLeft = pNewL; pRoot->pRight = pNewR; pNewL->pLeft = pNewLL; pNewL->pRight = pNewLR; pNewR->pLeft = pNewRL; pNewR->pRight = pNewRR; return pRoot; } void main() { bstNode* pRoot = buildupTree(); std::cout<<medianInBST(pRoot)<<std::endl; system("pause"); }

說實話這樣的題目還是比較喜歡的，考到了很多的概念和想法，

a.中序遍歷

b.BST轉化成DLL

c.尋找連結串列的中間節點

如果任何一個問題割裂開了問， 都是比較容易解決的。困難就在於如何用已知的辦法組合地解決未知的問題，發人深思，餘是以記之。