根据先序和中序序列重建二叉树

#include "stdafx.h"
#include <iostream>
#include <exception>
#include <stack>
using namespace std;

/*
     重建二叉树
题目：输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树.
     假设输入的前序遍历和中序遍历的结果中都不含重复的数字.例如
     输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6},
     则重建出图所示的二叉树并输出它的头结点。二叉树结点的定义如下:
*/

struct BinaryTreeNode
{
    int  m_nValue;
    BinaryTreeNode* m_pLeft;
    BinaryTreeNode* m_PRight;
};

BinaryTreeNode* ConstructConre(int* startPreorder,int* endPreorder,int* startInorder,int* endInorder)
{
    int rootValue = startPreorder[0];
    BinaryTreeNode* root = new BinaryTreeNode();
    root->m_nValue = rootValue;
    root->m_pLeft = root->m_PRight = NULL;
    if(startPreorder == endPreorder)
    {
        if(startInorder == endInorder && *startPreorder ==*startInorder)
            return root;
        else
            throw std::exception("invalid input");
    }
    //在中序遍历中找到根结点的值
    int* rootInorder = startInorder;
    while(rootInorder <= endInorder&& *rootInorder !=rootValue)
        ++ rootInorder;
    if(rootInorder ==endInorder && *rootInorder != rootValue)
        throw std::exception("Invalid input.");
    int leftLength = rootInorder - startInorder;
    int *leftPreorderEnd = startPreorder +leftLength;
    if(leftLength >0)
    {
        //构建左子树
        root->m_pLeft = ConstructConre(startPreorder+1,leftPreorderEnd,startInorder,rootInorder -1);
    }
    if(leftLength<endPreorder - startPreorder)
    {
        //构建右子树
        root->m_PRight = ConstructConre(leftPreorderEnd +1,endPreorder,rootInorder+1,endInorder);
    }
    return root;
}

BinaryTreeNode* Construct(int *preOrder,int* inOrder,int length)
{
    if(preOrder==NULL||inOrder==NULL||length<=0)
    {
        return NULL;
    }
    return ConstructConre(preOrder,preOrder+length-1,inOrder,inOrder+length-1);
}
int _tmain(int argc, _TCHAR* argv[])
{ 
    return 0 ;
}