[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;}}
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