已知前序和中序遍历恢复二叉树（递归）

#include<iostream>
using namespace std;
#define TREELEN  6
//数据结构定义
struct NODE
{
NODE* pLeft;         //左子树
NODE* pRight;        //右子树
char chValue;        //该节点的值
};

void ReBuild(char* pPreOrder,char* pInOrder,int nTreeLen,NODE** pRoot)
{
//检查边界条件
if(pPreOrder==NULL || pInOrder==NULL)
{
return;
}

//获得前序遍历的第一个节点
NODE* pTemp = new NODE;
pTemp->chValue = *pPreOrder;
pTemp->pLeft   = NULL;
pTemp->pRight  = NULL;

//如果节点为空，把当前节点复制到根节点
if(*pRoot == NULL)
{
*pRoot = pTemp;
}

//如果当前树长度为1，那么已经是最后一个节点
if(nTreeLen == 1)
{
return;
}

//寻找子树长度
char* pOrgInOrder = pInOrder;
char* pLeftEnd = pInOrder;
int nTempLen = 0;

//找到左子树的结尾
while(*pPreOrder != *pLeftEnd)
{
if(pPreOrder==NULL || pLeftEnd==NULL)
{
return;
}
nTempLen++;
//记录临时长度，以免溢出
if(nTempLen > nTreeLen)
{
break;
}
pLeftEnd++;
}

//寻找左子树长度
int nLeftLen = 0;
nLeftLen = (int)(pLeftEnd-pOrgInOrder);

//寻找右子树长度
int nRightLen = 0;
nRightLen = nTreeLen - nLeftLen - 1;

//重建左子树
if(nLeftLen > 0)
{
ReBuild(pPreOrder+1,pInOrder,nLeftLen,&((*pRoot)->pLeft));
}

//重建右子树
if(nRightLen > 0)
{
ReBuild(pPreOrder+nLeftLen+1,pInOrder+nLeftLen+1,nRightLen,&((*pRoot)->pRight));
}

}

//前序遍历结果
void PrePrint(NODE* pRoot)
{
if(pRoot == NULL)
{
return;
}
cout<<pRoot->chValue<<" ";
PrePrint(pRoot->pLeft);
PrePrint(pRoot->pRight);
}

//中序遍历结果
void InPrint(NODE* pRoot)
{
if(pRoot == NULL)
{
return;
}
InPrint(pRoot->pLeft);
cout<<pRoot->chValue<<" ";
InPrint(pRoot->pRight);
}

void main()
{
char szPreOrder[TREELEN] = {'a','b','d','c','e','f'};
char szInOrder[TREELEN]  = {'d','b','a','e','c','f'};
NODE* pRoot = NULL;
ReBuild(szPreOrder,szInOrder,TREELEN,&pRoot);
PrePrint(pRoot);
cout<<endl<<endl;;

InPrint(pRoot);
cout<<endl;
}

/*
a b d c e f

d b a e c f
*/