输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表

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微软笔试题：

输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。

要求不能创建任何新的结点，只调整指针的指向。

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

转换成双向链表
4=6=8=10=12=14=16。

可分如下步骤来实现：

1）定义二叉树节点

typedef struct BSTreeNode{int value;struct BSTreeNode *lchild;struct BSTreeNode *rchild;}BSTreeNode;

2）建立二分查找树，二分查找树的建立流程图如下：

其实现的源代码如下：

//Create the binary treevoid addBSTreeNode(BSTreeNode* pCurrent, int value){if(NULL == pCurrent){BSTreeNode *pBSTree = (BSTreeNode*)malloc(sizeof(BSTreeNode));pBSTree->lchild = NULL;pBSTree->rchild = NULL;pBSTree->value = value;pCurrent = pBSTree;}else{if((pCurrent->value) > value){addBSTreeNode(pCurrent->lchild, value);}else if((pCurrent->value) < value){addBSTreeNode(pCurrent->rchild, value);}else{printf("Wrong, Repeated insert the node");}}}

3）对建立的二叉查找树进行到双向链表的转换

源代码如下：

// convert binary tree to list.void convertToDoubleList(BSTreeNode *pCurrent){pCurrent->lchild = ListIndex;if(NULL != ListIndex){ListIndex->rchild = pCurrent;}else{Head = pCurrent;}ListIndex = pCurrent;printf("%d\n",pCurrent->value);}// ergodic the binary tree using inorder.void ergodicBSTree(BSTreeNode *pCurrent){if(NULL == pCurrent){return;}if(NULL != pCurrent->lchild){ergodicBSTree(pCurrent->lchild);}convertToDoubleList(pCurrent);if(NULL != pCurrent->rchild){ergodicBSTree(pCurrent->rchild);}}

4）主函数如下：

int main(){BSTreeNode *pRoot = NULL;ListIndex = NULL;Head = NULL;addBSTreeNode(pRoot, 10);addBSTreeNode(pRoot, 4);addBSTreeNode(pRoot, 6);addBSTreeNode(pRoot, 8);addBSTreeNode(pRoot, 12);addBSTreeNode(pRoot, 14);addBSTreeNode(pRoot, 15);addBSTreeNode(pRoot, 16);ergodicBSTree(pRoot);system("pause");return 0;}

注：程序有点问题，需要修改