[Microsoft面试]把二元查找树转变成排序的双向链表

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。首先我们定义的二元查找树节点的数据结构如下：struct BSTreeNode{int m_nValue; // value of nodeBSTreeNode *m_pLeft; // left child of nodeBSTreeNode *m_pRight; // right child of node};

This isa traditional problem that can be solved using recursion.For each node, connect the double linked lists created from left and rightchild node to form a full list./** * @param root The root node of the tree * @return The head node of the converted list. */BSTreeNode * treeToLinkedList(BSTreeNode * root) {  BSTreeNode * head, * tail;  helper(head, tail, root);  return head;}voidhelper(BSTreeNode *& head, BSTreeNode *& tail, BSTreeNode *root) {  BSTreeNode *lt, *rh;  if (root == NULL) {    head = NULL, tail = NULL;    return;  }  helper(head, lt, root->m_pLeft);  helper(rh, tail, root->m_pRight);  if (lt!=NULL) {    lt->m_pRight = root;    root->m_pLeft = lt;  } else  {    head = root;  }  if (rh!=NULL) {    root->m_pRight=rh;    rh->m_pLeft = root;  } else {    tail = root;  }}