精选微软数据结构算法面试100题

1.把二元查找树转变成排序的双向链表 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 node BSTreeNode *m_pLeft; // left child of node BSTreeNode *m_pRight; // right child of node }; #include <stdio.h> #include <iostream.h> struct BSTreeNode { int m_nValue; // value of node BSTreeNode *m_pLeft; // left child of node BSTreeNode *m_pRight; // right child of node }; typedef BSTreeNode DoubleList; DoubleList * pHead; DoubleList * pListIndex; void convertToDoubleList(BSTreeNode * pCurrent); // 创建二元查找树 void addBSTreeNode(BSTreeNode * & pCurrent, int value) { if (NULL == pCurrent) { BSTreeNode * pBSTree = new BSTreeNode(); pBSTree->m_pLeft = NULL; pBSTree->m_pRight = NULL; pBSTree->m_nValue = value; pCurrent = pBSTree; } else { if ((pCurrent->m_nValue) > value) { addBSTreeNode(pCurrent->m_pLeft, value); } else if ((pCurrent->m_nValue) < value) { addBSTreeNode(pCurrent->m_pRight, value); } else { //cout<<"重复加入节点"<<endl; } } } // 遍历二元查找树中序 void ergodicBSTree(BSTreeNode * pCurrent) { if (NULL == pCurrent) { return; } if (NULL != pCurrent->m_pLeft) { ergodicBSTree(pCurrent->m_pLeft); } // 节点接到链表尾部 convertToDoubleList(pCurrent); // 右子树为空 if (NULL != pCurrent->m_pRight) { ergodicBSTree(pCurrent->m_pRight); } } // 二叉树转换成list void convertToDoubleList(BSTreeNode * pCurrent) { pCurrent->m_pLeft = pListIndex; if (NULL != pListIndex) { pListIndex->m_pRight = pCurrent; } else { pHead = pCurrent; } pListIndex = pCurrent; cout<<pCurrent->m_nValue<<endl; } int main() { BSTreeNode * pRoot = NULL; pListIndex = NULL; pHead = 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); return 0; }