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 100
int path[N]={0};// used for calculating the path
using 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;
	}
	else
		if(root->m_nValue>value)
		{
			buildBSTree(root->m_pLeft,value);
		}
		else
		if(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 pop
			stk.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 pop
			t=p;
			p=t->m_pRight;// p serve the searching record in the tree
			if(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);
		}
		else
			return ;
	}
}
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;
}