二进制查找树转换为双向链表
全然依照海涛哥剑指offer里边的递归思路来写的,基本一样。仅作学习验证,努力锻炼,努力学习。
题目: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。
要求不能创建不论什么新的结点,仅仅调整指针的指向。
比方将二元查找树
10
/ \
6 14
/ \ / \
4 8 12 16
转换成双向链表
4=6=8=10=12=14=16。
code例如以下://Change a BSTree to a sorted double linklist struct BSTreeNode { int value; BSTreeNode *left; BSTreeNode *right; }head; //Create a node of BSTree for test BSTreeNode* CreateNode(int value) { BSTreeNode *node = new BSTreeNode(); node->value = value; node->left = NULL; node->right = NULL; return node; } //using a lastNode pointer to convert the pointer of the node void ConvertNode(BSTreeNode *head, BSTreeNode **lastNode) { if(head == NULL) return; BSTreeNode *pCurrent = head; if(pCurrent->left != NULL) ConvertNode(pCurrent->left, lastNode); pCurrent->left = *lastNode; if(*lastNode != NULL) (*lastNode)->right = pCurrent; *lastNode = pCurrent; if(pCurrent->right != NULL) ConvertNode(pCurrent->right, lastNode); } //lastNode pointer is pointing to the rear,so the head can be obtained BSTreeNode* Convert(BSTreeNode *head) { if(head == NULL) return NULL; BSTreeNode *lastNode = NULL; ConvertNode(head, &lastNode); while(lastNode->left != NULL) lastNode = lastNode->left; return lastNode; } //main function for test int main() { BSTreeNode *head = CreateNode(10); head->left = CreateNode(8); head->left->left = CreateNode(7); head->left->right = CreateNode(9); head->right = CreateNode(12); head->right->left = CreateNode(11); head = Convert(head); while(head != NULL) { printf("%d ",head->value); head = head->right; } }