【面试题027】二叉搜索树与双向链表
题目：
输入一棵二叉搜索树，将该二叉搜索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，只能调用书中结点的指向。
二叉树的结点定义如下：
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struct BinaryTreeNode
{ int m_nValue; BinaryTreeNode *m_pLeft; BinartTreeNode *m_pRight; } |
ConvertBinarySearchTree.cpp:
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#include <iostream>
#include "BinaryTree.h" using namespace std; void ConvertNode(BinaryTreeNode *pNode, BinaryTreeNode **pLastNodeInList); BinaryTreeNode *Convert(BinaryTreeNode *pRootOfTree) { BinaryTreeNode *pLastNodeInList = NULL; ConvertNode(pRootOfTree, &pLastNodeInList); // pLastNodeInList指向双向链表的尾结点， // 我们需要返回头结点 BinaryTreeNode *pHeadOfList = pLastNodeInList; while(pHeadOfList != NULL && pHeadOfList->m_pLeft != NULL) pHeadOfList = pHeadOfList->m_pLeft; return pHeadOfList; } void ConvertNode(BinaryTreeNode *pNode, BinaryTreeNode **pLastNodeInList) { if(pNode == NULL) return; BinaryTreeNode *pCurrent = pNode; if (pCurrent->m_pLeft != NULL) ConvertNode(pCurrent->m_pLeft, pLastNodeInList); pCurrent->m_pLeft = *pLastNodeInList; if(*pLastNodeInList != NULL) (*pLastNodeInList)->m_pRight = pCurrent; *pLastNodeInList = pCurrent; if (pCurrent->m_pRight != NULL) ConvertNode(pCurrent->m_pRight, pLastNodeInList); } // ====================测试代码==================== void PrintDoubleLinkedList(BinaryTreeNode *pHeadOfList) { BinaryTreeNode *pNode = pHeadOfList; printf("The nodes from left to right are: "); while(pNode != NULL) { printf("%d ", pNode->m_nValue); if(pNode->m_pRight == NULL) break; pNode = pNode->m_pRight; } printf(" The nodes from right to left are: "); while(pNode != NULL) { printf("%d ", pNode->m_nValue); if(pNode->m_pLeft == NULL) break; pNode = pNode->m_pLeft; } printf(" "); } void DestroyList(BinaryTreeNode *pHeadOfList) { BinaryTreeNode *pNode = pHeadOfList; while(pNode != NULL) { BinaryTreeNode *pNext = pNode->m_pRight; delete pNode; pNode = pNext; } } void Test(char *testName, BinaryTreeNode *pRootOfTree) { if(testName != NULL) printf("%s begins: ", testName); PrintTree(pRootOfTree); BinaryTreeNode *pHeadOfList = Convert(pRootOfTree); PrintDoubleLinkedList(pHeadOfList); } // 10 // / // 6 14 // / / // 4 8 12 16 void Test1() { BinaryTreeNode *pNode10 = CreateBinaryTreeNode(10); BinaryTreeNode *pNode6 = CreateBinaryTreeNode(6); BinaryTreeNode *pNode14 = CreateBinaryTreeNode(14); BinaryTreeNode *pNode4 = CreateBinaryTreeNode(4); BinaryTreeNode *pNode8 = CreateBinaryTreeNode(8); BinaryTreeNode *pNode12 = CreateBinaryTreeNode(12); BinaryTreeNode *pNode16 = CreateBinaryTreeNode(16); ConnectTreeNodes(pNode10, pNode6, pNode14); ConnectTreeNodes(pNode6, pNode4, pNode8); ConnectTreeNodes(pNode14, pNode12, pNode16); Test("Test1", pNode10); DestroyList(pNode4); } // 5 // / // 4 // / // 3 // / // 2 // / // 1 void Test2() { BinaryTreeNode *pNode5 = CreateBinaryTreeNode(5); BinaryTreeNode *pNode4 = CreateBinaryTreeNode(4); BinaryTreeNode *pNode3 = CreateBinaryTreeNode(3); BinaryTreeNode *pNode2 = CreateBinaryTreeNode(2); BinaryTreeNode *pNode1 = CreateBinaryTreeNode(1); ConnectTreeNodes(pNode5, pNode4, NULL); ConnectTreeNodes(pNode4, pNode3, NULL); ConnectTreeNodes(pNode3, pNode2, NULL); ConnectTreeNodes(pNode2, pNode1, NULL); Test("Test2", pNode5); DestroyList(pNode1); } // 1 // // 2 // // 3 // // 4 // // 5 void Test3() { BinaryTreeNode *pNode1 = CreateBinaryTreeNode(1); BinaryTreeNode *pNode2 = CreateBinaryTreeNode(2); BinaryTreeNode *pNode3 = CreateBinaryTreeNode(3); BinaryTreeNode *pNode4 = CreateBinaryTreeNode(4); BinaryTreeNode *pNode5 = CreateBinaryTreeNode(5); ConnectTreeNodes(pNode1, NULL, pNode2); ConnectTreeNodes(pNode2, NULL, pNode3); ConnectTreeNodes(pNode3, NULL, pNode4); ConnectTreeNodes(pNode4, NULL, pNode5); Test("Test3", pNode1); DestroyList(pNode1); } // 树中只有1个结点 void Test4() { BinaryTreeNode *pNode1 = CreateBinaryTreeNode(1); Test("Test4", pNode1); DestroyList(pNode1); } // 树中没有结点 void Test5() { Test("Test5", NULL); } int main() { Test1(); Test2(); Test3(); Test4(); Test5(); return 0; } |
BinaryTree.h:
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#ifndef _Binary_Tree_
#define _Binary_Tree_ struct BinaryTreeNode { int m_nValue; BinaryTreeNode *m_pLeft; BinaryTreeNode *m_pRight; }; BinaryTreeNode *CreateBinaryTreeNode(int value); void ConnectTreeNodes(BinaryTreeNode *pParent, BinaryTreeNode *pLeft, BinaryTreeNode *pRight); void PrintTreeNode(BinaryTreeNode *pNode); void PrintTree(BinaryTreeNode *pRoot); void DestroyTree(BinaryTreeNode *pRoot); #endif /*_Binary_Tree_*/ |
BinaryTree.cpp:
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#include "BinaryTree.h"
#include <cstdio> BinaryTreeNode *CreateBinaryTreeNode(int value) { BinaryTreeNode *pNode = new BinaryTreeNode(); pNode->m_nValue = value; pNode->m_pLeft = NULL; pNode->m_pRight = NULL; return pNode; } void ConnectTreeNodes(BinaryTreeNode *pParent, BinaryTreeNode *pLeft, BinaryTreeNode *pRight) { if(pParent != NULL) { pParent->m_pLeft = pLeft; pParent->m_pRight = pRight; } } void PrintTreeNode(BinaryTreeNode *pNode) { if(pNode != NULL) { printf("value of this node is: %d ", pNode->m_nValue); if(pNode->m_pLeft != NULL) printf("value of its left child is: %d. ", pNode->m_pLeft->m_nValue); else printf("left child is null. "); if(pNode->m_pRight != NULL) printf("value of its right child is: %d. ", pNode->m_pRight->m_nValue); else printf("right child is null. "); } else { printf("this node is null. "); } printf(" "); } void PrintTree(BinaryTreeNode *pRoot) { PrintTreeNode(pRoot); if(pRoot != NULL) { if(pRoot->m_pLeft != NULL) PrintTree(pRoot->m_pLeft); if(pRoot->m_pRight != NULL) PrintTree(pRoot->m_pRight); } } void DestroyTree(BinaryTreeNode *pRoot) { if(pRoot != NULL) { BinaryTreeNode *pLeft = pRoot->m_pLeft; BinaryTreeNode *pRight = pRoot->m_pRight; delete pRoot; pRoot = NULL; DestroyTree(pLeft); DestroyTree(pRight); } } |
运算结果：
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Test1 begins:
value of this node is: 10 value of its left child is: 6. value of its right child is: 14. value of this node is: 6 value of its left child is: 4. value of its right child is: 8. value of this node is: 4 left child is null. right child is null. value of this node is: 8 left child is null. right child is null. value of this node is: 14 value of its left child is: 12. value of its right child is: 16. value of this node is: 12 left child is null. right child is null. value of this node is: 16 left child is null. right child is null. The nodes from left to right are: 4 6 8 10 12 14 16 The nodes from right to left are: 16 14 12 10 8 6 4 Test2 begins: value of this node is: 5 value of its left child is: 4. right child is null. value of this node is: 4 value of its left child is: 3. right child is null. value of this node is: 3 value of its left child is: 2. right child is null. value of this node is: 2 value of its left child is: 1. right child is null. value of this node is: 1 left child is null. right child is null. The nodes from left to right are: 1 2 3 4 5 The nodes from right to left are: 5 4 3 2 1 Test3 begins: value of this node is: 1 left child is null. value of its right child is: 2. value of this node is: 2 left child is null. value of its right child is: 3. value of this node is: 3 left child is null. value of its right child is: 4. value of this node is: 4 left child is null. value of its right child is: 5. value of this node is: 5 left child is null. right child is null. The nodes from left to right are: 1 2 3 4 5 The nodes from right to left are: 5 4 3 2 1 Test4 begins: value of this node is: 1 left child is null. right child is null. The nodes from left to right are: 1 The nodes from right to left are: 1 Test5 begins: this node is null. The nodes from left to right are: The nodes from right to left are: 请按任意键继续. . . |