剑指offer 面试题6

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}，则重建出二叉树并输入头结点。
二叉树定义如下：
struct BinaryNode
{
int m_nValue;
BinaryNode * m_pLeft;
BinaryNode * m_pRight;
};

本题考查中序遍历和前序遍历之间的联系，并且考察面试者的编程水准。

#include<iostream>using namespace std;struct BinaryNode{    int m_nValue;    BinaryNode *m_pLeft;    BinaryNode *m_pRight;};BinaryNode * ConCore(int *beginPreOrder,int *endPreOrder,int *beginInOrder,int *endInOrder){    BinaryNode *pRoot=NULL;    int rootValue = beginPreOrder[0];    pRoot = new BinaryNode();    pRoot->m_nValue = rootValue;    pRoot->m_pLeft = pRoot->m_pRight = NULL;    if (beginPreOrder == endPreOrder)    {        if (beginInOrder == endInOrder && *beginPreOrder == *beginInOrder)        {            return pRoot;        }        else            throw exception("Invalid input");    }    int *InOrderPos = beginInOrder;    while (InOrderPos<=endInOrder && *InOrderPos!=rootValue)    {        ++InOrderPos;    }    if (InOrderPos > endInOrder)        throw exception("Invalid input");    int LeftLegth = InOrderPos - beginInOrder;    if (LeftLegth > 0)        pRoot->m_pLeft = ConCore(beginPreOrder + 1, beginPreOrder + LeftLegth,beginInOrder,InOrderPos-1);    if (LeftLegth < endInOrder - beginInOrder)        pRoot->m_pRight = ConCore(beginPreOrder+LeftLegth+1,endPreOrder,InOrderPos+1,endInOrder);    return pRoot;}BinaryNode * Con(int *preOrder,int *InOrder,int length){    if (NULL == preOrder || NULL == InOrder || length <= 0)        return NULL;    return ConCore(preOrder,preOrder+length-1,InOrder,InOrder+length-1);}void PostOrder(BinaryNode *head){    if (head == NULL)        return;    PostOrder(head->m_pLeft);    PostOrder(head->m_pRight);    cout << head->m_nValue << " ";}int main(){    int p1[] = {1,2,4,7,3,5,6,8};    int p2[] = {4,7,2,1,5,3,8,6};    BinaryNode *head = Con(p1,p2,8);    PostOrder(head);    return 0;}