算法--根据二叉树前序和中序遍历序列，求后续遍历序列

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根据二叉树前序和中序遍历序列，求后续遍历序列。

#include<stdio.h>#include<string.h>#include<stdlib.h>//根据字符分割字符串void SplitSeq1(const char* srcSeq, char splitor, char* &left, char* &right){int len=strlen(srcSeq);int i=0;for(i=0; i<len; ++i){if(srcSeq[i]==splitor)break;}if(i==len){left=right=NULL;return;}left=(char*)malloc(sizeof(char)*(i+1));right=(char*)malloc(sizeof(char)*(len-i));int pos=i;for(i=0; i<pos; i++) left[i]=srcSeq[i];left[i]=0;for(i=pos+1; i<len; i++) right[i-pos-1]=srcSeq[i];right[len-pos-1]=0;}//分割字符串2void SplitSeq2(const char* srcStr, int len1, int len2, char* &left, char* &right){left=(char*)malloc(sizeof(char)*(len1+1));right=(char*)malloc(sizeof(char)*(len2+1));int pos=1, i=0;for( i=0; i<len1; i++)left[i]=srcStr[pos+i];left[i]=0;pos=pos+i;for(i=0; i<len2; i++)right[i]=srcStr[pos+i];right[i]=0;}void PrintPostSeq(const char* preSeq, const char* midSeq){if(*preSeq==0 || *midSeq==0) return;if(*midSeq==*preSeq && strlen(preSeq)==1){printf("%c", *preSeq);  return;}char node=*preSeq;char *leftSeqMid=NULL, *rightSeqMid=NULL;SplitSeq1(midSeq, node, leftSeqMid, rightSeqMid);int len1=strlen(leftSeqMid);   int len2=strlen(rightSeqMid);char *leftSeqPre=NULL, *rightSeqPre=NULL;SplitSeq2(preSeq, len1, len2, leftSeqPre, rightSeqPre);PrintPostSeq(leftSeqPre, leftSeqMid);PrintPostSeq(rightSeqPre, rightSeqMid);free(leftSeqPre); free(leftSeqMid);free(rightSeqPre); free(rightSeqMid); printf("%c", node);}int main(){char strPre[]="BCAD", strMid[]="CBAD";PrintPostSeq(strPre, strMid);return 0;}

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