重建二叉树

来源：互联网发布：php 无刷新post 编辑：程序博客网时间：2024/05/01 01:33

#include <stdio.h>
#include <stdlib.h>
#define TREELEN 6

struct Node
{
    struct Node * pLeft;
    struct Node * pRight;
    char chValue;
};

void ReBuildTree(char * pPreOrder, char * pInOrder, int nTreeLen, struct 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
    struct Node * pTemp = (struct Node *)malloc(sizeof(struct 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;

    //找到左子树德结尾
    while(*pPreOrder != *pLeftEnd)
    {
        if(pPreOrder == NULL || pLeftEnd == NULL)
        {
            return;
        }

nTempLen++;

        //记录临时表长度，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)
    {
        ReBuildTree(pPreOrder + 1, pInOrder, nLeftLen, &((*pRoot) -> pLeft));
    }

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

}

int main()
{
    char szPreOrder[TREELEN] = {'a', 'b', 'd', 'c', 'e', 'f'};
    char szInOrder[TREELEN] = {'d', 'b', 'a', 'e', 'c', 'f'};
    struct Node * pRoot = NULL;
    ReBuildTree(szPreOrder, szInOrder, TREELEN, &pRoot);
    //printf("Hello world!\n");

return 0;
}