【编程之美】3.9重建二叉树

/*
start time = 15:55
end time = 16:23
*/
#include<iostream>
using namespace std;

typedef struct NODE{
    NODE *pLeft, *pRight;
    char chValue;
}NODE;

void Rebuild(char* pPreOrder, char* pInOrder, int nTreeLen, NODE** pRoot)
{
    if(pPreOrder == NULL || pInOrder == NULL) //边界条件不要忘了 这是看答案后补上的
    {
        return;
    }
    if(nTreeLen <= 0)
    {
        return;
    }
    //重建根节点 一定是先序遍历的第一个值
    *pRoot = new NODE;
    (*pRoot)->chValue = pPreOrder[0];
    (*pRoot)->pLeft = NULL;
    (*pRoot)->pRight = NULL;

    //查找根在中序遍历中的位置，把中序遍历表转化为左子树和右子树两个部分
    int iLocateRoot;
    for(iLocateRoot = 0; iLocateRoot < nTreeLen; iLocateRoot++)
    {
        if(pInOrder[iLocateRoot] == pPreOrder[0])
        {
            break;
        }
    }

    //重建左子树
    Rebuild(pPreOrder+1, pInOrder, iLocateRoot, &((*pRoot)->pLeft));
    //重建右子树
    Rebuild(pPreOrder+iLocateRoot+1, pInOrder + iLocateRoot + 1, nTreeLen - iLocateRoot - 1, &((*pRoot)->pRight));
}

int main()
{
    char pre[7] = {‘a‘,‘b‘,‘d‘,‘c‘,‘e‘,‘f‘,‘\0‘};
    char in[7] = {‘d‘,‘b‘,‘a‘,‘e‘,‘c‘,‘f‘,‘\0‘};
    NODE * T = NULL;
    Rebuild(pre, in, 7, &T);

    return 0;
}