// 程序员面试题精选100题(48)-二叉树两结点的最低共同父结点.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>#include <list>using namespace std;#define N 7struct TNode{char chValue;TNode *leftChild;TNode *rightChild;};struct ListNode {//used for calculating the lowest common root,two lists record the search routes of the two point to be searchedchar chValue;ListNode* next;};TNode* rebuildTree(char *preOrder,int start1,char *midOrder,int start2,int n)//根据前序和中序建立二叉树 长度为N{TNode *root;if(n==0) return NULL;root = new TNode;int pivet;char chroot;chroot=preOrder[start1];root->chValue = chroot;for(int i=start2;i<start2+n;i++)//include start1 {if(midOrder[i]==chroot) { pivet = i;break;}}if(pivet-start2>0)root->leftChild=rebuildTree(preOrder,start1+1,midOrder,start2,pivet-start2);else root->leftChild = NULL;if(n-pivet+start2-1>0)//pivet-start2 is the half number,root->rightChild=rebuildTree(preOrder,start1+1+pivet-start2,midOrder,pivet+1,n-pivet+start2-1);else root->rightChild = NULL;return root;}void printList(ListNode* head){ListNode* temp=head;cout<<" list is ";while(temp!=NULL){cout<<temp->chValue<<" ";temp=temp->next;}cout<<endl;}bool SearchWithList(TNode* root,char val,ListNode* head)// record the searching route in the list{if (root==NULL)//does not find{return false;}if (root->chValue==val){/*ListNode *temp=new ListNode;//here, if this is none, then the same as the answertemp->chValue = root->chValue;temp->next=NULL;head->next=temp;*/return true;}else{ListNode *temp1=new ListNode;temp1->chValue = root->chValue;temp1->next=NULL;head->next=temp1;if (!SearchWithList(root->leftChild,val,temp1))// if there no exist in the left node, then turn to the right node,otherwise, stop{ListNode *temp2=new ListNode;temp2->chValue = root->chValue;temp2->next=NULL;head->next=temp2;//change head->next, previous become invalidateif (!SearchWithList(root->rightChild,val,temp2)){head->next=NULL;return false;}elsereturn true;}elsereturn true;}}char LowestCommonRoot(TNode* root,char ch1,char ch2){ListNode *head1,*head2;head1= new ListNode;head1->chValue = '#';// head value is spacehead1->next=NULL;head2= new ListNode;head2->chValue = '#';head2->next=NULL;bool bl1=SearchWithList(root,ch1,head1);bool bl2=SearchWithList(root,ch2,head2);if (bl1){printList(head1);}if (bl2){printList(head2);}if (!bl1&&!bl2){return '#';//means does not exist}else{ListNode *p1=head1,*p2=head2;if (p1->chValue!=p2->chValue){return '#';}while(p1->next!=NULL&&p2->next!=NULL&&p1->next->chValue==p2->next->chValue){p1=p1->next;p2=p2->next;}return p1->chValue;}}//////////////////////////////////////////////////////////////////////////bool GetNodePath(TNode* pHead, TNode* pNode, std::list<TNode*>& path){if(pHead == pNode)return true;path.push_back(pHead);bool found = false;if(pHead->leftChild != NULL)found = GetNodePath(pHead->leftChild, pNode, path);if(!found && pHead->rightChild)found = GetNodePath(pHead->rightChild, pNode, path);if(!found)path.pop_back();return found;}TNode* LastCommonNode(const std::list<TNode*>& path1,const std::list<TNode*>& path2){std::list<TNode*>::const_iterator iterator1 = path1.begin();std::list<TNode*>::const_iterator iterator2 = path2.begin();TNode* pLast = NULL;while(iterator1 != path1.end() && iterator2 != path2.end()){if(*iterator1 == *iterator2)pLast = *iterator1;iterator1++;iterator2++;}return pLast;}TNode* LastCommonParent_2(TNode* pHead, TNode* pNode1, TNode* pNode2){if(pHead == NULL || pNode1 == NULL || pNode2 == NULL)return NULL;std::list<TNode*> path1;GetNodePath(pHead, pNode1, path1);std::list<TNode*> path2;GetNodePath(pHead, pNode2, path2);return LastCommonNode(path1, path2);}//////////////////////////////////////////////////////////////////////////int _tmain(int argc, _TCHAR* argv[]){char preOrder[N]={'A','B','C','D','E','F','G'},midOrder[N]={'C','B','D','A','F','E','G'};/*cout<<"input preorder"<<endl;for(int i=0;i<N;i++)cin>>preOrder[i];cout<<"input midorder"<<endl;for(int i=0;i<N;i++)cin>>midOrder[i];*/TNode *root;root=rebuildTree(preOrder,0,midOrder,0,N);cout<<endl<<" root value is "<<root->chValue<<endl;cout<<endl<<"common root is "<<LowestCommonRoot(root,'F','E')<<endl;TNode *lnode,*rnode;//lnode=root->leftChild; // result is A? right //rnode=root->leftChild->rightChild;lnode=root->rightChild;//->leftChild;rnode=root->rightChild->rightChild;cout<<lnode->chValue<<endl;cout<<rnode->chValue<<endl;cout<<" the second method value is "<<LastCommonParent_2(root,lnode,rnode)->chValue<<endl;system("pause");return 0;}what are the differences between the stardard answer and yours? there are many aspects to learn.
