微软面试100题之一，之四 二叉查找树变双向链表 和为某一值的所有路径

// 微软100题之一二叉查找树变双向链表.cpp : 定义控制台应用程序的入口点。//微软100题之四 二叉查找树找出和为某一值的所有路径#include "stdafx.h"#include <iostream>#include <stack>#define N 100int path[N]={0};// used for calculating the pathusing namespace std;struct BSTreeNode{int m_nValue;BSTreeNode *m_pLeft;BSTreeNode *m_pRight;};void buildBSTree(BSTreeNode *&root, int value){if(root==NULL){root = new BSTreeNode();root->m_nValue = value;root->m_pLeft = NULL;root->m_pRight = NULL;}elseif(root->m_nValue>value){buildBSTree(root->m_pLeft,value);}elseif(root->m_nValue<value)// insert into rightchild{buildBSTree(root->m_pRight,value);}else ;//when exist same element skip}void midprint(BSTreeNode *root)//要传递指针的引用才好使。 unreverse mid 每一个元素都要入栈，出战的顺序即是遍历的顺序{stack<BSTreeNode*> stk;BSTreeNode *p=root;while(p!=NULL||stk.empty()==0){while(p!=NULL){stk.push(p);p=p->m_pLeft;}if(stk.empty()==0){p=stk.top();cout<<p->m_nValue<<" ";//all the element should be push into the stack then popstk.pop();p=p->m_pRight;}}}BSTreeNode *treeToList(BSTreeNode *root)// 在中序非递归遍历时处理每个元素。每一个元素都要入栈，出战的顺序即是遍历的顺序{stack<BSTreeNode*> stk;BSTreeNode *p=root,*q=NULL,*t,*head;//q initialized as NULL while(p!=NULL||stk.empty()==0){while(p!=NULL){stk.push(p);p=p->m_pLeft;}if(stk.empty()==0){p=stk.top();//cout<<p->m_nValue<<" ";//all the element should be push into the stack then popt=p;p=t->m_pRight;// p serve the searching record in the treeif(q==NULL){q=t;head=q;q->m_pRight=NULL;q->m_pLeft=q;// the leftchild of the head refers to itself}else{t->m_pRight = q->m_pRight;q->m_pRight = t;t->m_pLeft = q;q = t;}stk.pop();}}return head;}void searchList(BSTreeNode *root){BSTreeNode *p=root;while(p!=NULL){cout<<p->m_nValue<<" ";p=p->m_pRight;}}void calculatePath(BSTreeNode *root,int remain,int i)// i represent the layer{if(root->m_nValue==remain)// find a path, from 0 layer to i layer{path[i]=remain;for(int k=0;k<=i;k++)cout<<path[k]<<" ";cout<<endl;}else if(root->m_nValue>remain)return ;else{if(root->m_pLeft!=NULL||root->m_pRight!=NULL){path[i]=root->m_nValue;if(root->m_pLeft!=NULL)calculatePath(root->m_pLeft,remain-root->m_nValue,i+1);if(root->m_pRight!=NULL)calculatePath(root->m_pRight,remain-root->m_nValue,i+1);}elsereturn ;}}int _tmain(int argc, _TCHAR* argv[]){BSTreeNode *root=NULL,*head;int n,m;cout<<" input n "<<endl;cin>>n;cout<<" input values "<<endl;int *value = new int[n];for(int i=0; i<n; i++)    {                   cin >> value[i];    }for (int i = 0; i < n; i++)   {    buildBSTree(root, value[i]);    }cout<<" search tree "<<endl;midprint(root);cout<<endl<<"find the path which sum is m, please input m "<<endl;cin>>m;calculatePath(root,m,0);cout<<endl<<" search list "<<endl;head = treeToList(root);searchList(head);system("pause");return 0;}