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<iostream> 15 16 #include<fstream> 17 18 #include<string> 19 20 using namespace std; 21 22 23 string prestr,midstr; 24 25 struct node 26 { 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<<p->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<midstr.length();i++) 82 { 83 if(midstr[i]==prestr[0]) 84 mark=i; 85 } 86 groot->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<<endl; 91 92 93 return 0; 94 95 96 }
留个影
USER: hubiao cave [cavehub1] TASK: heritage LANG: C++ Compiling... Compile: OK Executing... 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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