poj2255Tree Recovery【二叉树重构】

大意：告诉你一棵二叉树的先序遍历和中序遍历求该二叉树的后续遍历

 

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int maxn = 30;
 7 char pr[maxn], mi[maxn];
 8 
 9 struct Node {
10     int left;
11     int right;
12 }tr[maxn];
13 
14 int root;
15 
16 void init(int p) {
17     tr[p].left = -1;
18     tr[p].right = -1;
19 }
20 
21 void fason(int pa, int son) {
22     if(son < pa) {
23         if(tr[pa].left == -1) {
24             init(son);
25             tr[pa].left = son;
26         } else fason(tr[pa].left, son);
27     } else {
28         if(tr[pa].right == -1) {
29             init(son);
30             tr[pa].right = son;
31         } else fason(tr[pa].right, son);
32     }
33 }
34 
35 void build_tree() {
36     for(int i = 0; mi[i]; i++) {
37         if(mi[i] == pr[0]) {
38             root = i;
39             init(root);
40             break;
41         }
42     }
43     for(int i =1; pr[i]; i++) {
44         for(int j = 0; mi[j]; j++) {
45             if(mi[j] == pr[i]) {
46                 fason(root, j);
47                 break;
48             }
49         }
50     }
51 }
52 void pre_out(int p) {
53     printf("%c", mi[p]);
54     if(tr[p].left != -1) pre_out(tr[p].left);
55     if(tr[p].right != -1) pre_out(tr[p].right);
56 }
57 void mid_out(int p) {
58     if(tr[p].left != -1) mid_out(tr[p].left);
59     printf("%c", mi[p]);
60     if(tr[p].right != -1) mid_out(tr[p].right);
61 }
62 
63 void back_out(int p) {
64     if(tr[p].left != -1) back_out(tr[p].left);
65     if(tr[p].right != -1) back_out(tr[p].right);
66     printf("%c", mi[p]);
67 }
68 
69 int main() {
70     while(EOF != scanf("%s %s",pr, mi) ) {
71         build_tree();
72         pre_out(root);
73         puts("");
74         mid_out(root);
75         puts("");
76         back_out(root);
77         puts("");
78     }
79     return 0;
80 }
View Code

 

posted @ 2014-09-23 20:44  悠悠我心。  阅读(186)  评论(0)  编辑  收藏  举报