已知前序和中序，求后序遍历

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通过简单例子可以分析出，先由前序找出当前根，再通过中序找出左右子树序列，然后递归输出左右子树，最后输出根

#include "stdafx.h"#include <iostream>using namespace std;void GetEnOrder(char* startPreOrder,char* endPreOrder,char* startInOrder,char* endInOrder);void GetAll(char* preOrder,char* inOrder,int lenght){if (preOrder==NULL||inOrder==NULL||lenght<=0){return;}GetEnOrder(preOrder,preOrder+lenght-1,inOrder,inOrder+lenght-1);}void GetEnOrder(char* startPreOrder,char* endPreOrder,char* startInOrder,char* endInOrder){if (startPreOrder<=endPreOrder){char root=*startPreOrder;char* rootInOrder=startInOrder;while(rootInOrder<=endInOrder&&*rootInOrder!=root){rootInOrder++;}if (rootInOrder==endInOrder+1){throw std::exception("序列错误！");}int leftLen=rootInOrder-startInOrder;int rightLen=endInOrder-rootInOrder;if (leftLen>0){GetEnOrder(startPreOrder+1,startPreOrder+leftLen,startInOrder,rootInOrder-1);}if(rightLen>0){GetEnOrder(startPreOrder+leftLen+1,endPreOrder,rootInOrder+1,endInOrder);}cout<<root;}}int _tmain(int argc, _TCHAR* argv[]){char preOrder[]={'A','B','D','C'};char inOrder[]={'B','D','A','C'};GetAll(preOrder,inOrder,4);return 0;}