就是根据二叉树的中序，前序，求后序，我想法是先求出二叉树再后序出来。开始感觉想起来很混乱，后来拿笔在纸上写一写就清楚很多了

这题不是很难，思路要清晰即可

1 /* 2  3 ID: hubiao cave 4  5 PROG: heritage 6  7 LANG: C++ 8  9 */10 11 12 13 14 #include
     
      15 16 #include
      
       17 18 #include
       
        19 20 using namespace std;21 22 23 string prestr,midstr;24 25 struct node26 {27     char m;28     node* left,*right;29     node()30     {31         left=right=NULL;32     }33 };34 35 node* groot;36  ofstream fout("heritage.out");37 node* GetChild(int ms,int me,int ps,int pe)38 {39     if(ms>me||ps>pe)40         return NULL;41     char parentch=prestr[ps];42     43     int mark;44     for(int i=ms;i<=me;i++)45         if(midstr[i]==parentch)46     {47         mark=i;48         break;49     }50 51     node* pnode=new node;52     pnode->m=parentch;53     pnode->left=GetChild(ms,mark-1,ps+1,ps+mark-ms);54     pnode->right=GetChild(mark+1,me,ps+mark-ms+1,pe);55     return pnode;56 57 }58 59 void AfterPrint(node* p)60 {61     if(p->left)62         AfterPrint(p->left);63     if(p->right)64         AfterPrint(p->right);65     fout<
        
         m;66 }67 68 69 int main()70 71 {72 73     ifstream fin("heritage.in");74 75    76     77 fin>>midstr>>prestr;78     groot=new node;79     groot->m=prestr[0];80     int mark;81     for(int i=0;i
         
          left=GetChild(0,mark-1,1,mark);87     groot->right=GetChild(mark+1,midstr.length()-1,1+mark,prestr.length()-1);88 89     AfterPrint(groot);90     fout<
         
        
       
      
     

留个影

USER: hubiao cave [cavehub1]TASK: heritageLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3500 KB]   Test 2: TEST OK [0.000 secs, 3500 KB]   Test 3: TEST OK [0.000 secs, 3500 KB]   Test 4: TEST OK [0.000 secs, 3500 KB]   Test 5: TEST OK [0.000 secs, 3500 KB]   Test 6: TEST OK [0.000 secs, 3500 KB]   Test 7: TEST OK [0.000 secs, 3500 KB]   Test 8: TEST OK [0.000 secs, 3500 KB]   Test 9: TEST OK [0.000 secs, 3500 KB]All tests OK.

YOUR PROGRAM ('heritage') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.

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