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853天前#include<bits/stdc++.h> |
#define Tree int |
#define ElemType char |
using namespace std; |
struct TreeNode |
{ |
ElemType element; |
Tree left; |
Tree right; |
}T1[10],T2[10]; |
int TreeBuild(struct TreeNode T[]) |
{ |
int N, i; |
char l, r; |
cin>> N; |
if (N) |
{ |
int flag[N]; |
for ( i = 0 ; i < N ; i ++) |
flag[i] = 1; |
for ( i = 0 ; i < N ; i ++) |
{ |
cin>>T[i].element>>l>>r; |
if (l == '-') |
T[i].left = -1; |
else |
{ |
T[i].left = l - '0'; |
flag[T[i].left] = 0; |
} |
if (r == '-') |
T[i].right = -1; |
else |
{ |
T[i].right = r - '0'; |
flag[T[i].right] = 0; |
} |
} |
for ( i = 0 ; i < N ; i ++) |
{ |
if (flag[i]) |
return i; |
} |
} |
else |
return -1; |
} |
int judge(Tree R1, Tree R2) |
{ |
if ((R1 == -1) && (R2 == -1))//都是空树 |
return 1; |
if ( ((R1 == -1)&&( R2 != -1)) || (R1 == -1) && ( R2 != -1))//一个是空树，一个不是空树 |
return 0; |
if (T1[R1].element != T2[R2].element)//根结点元素不一样 |
return 0; |
if ((T1[R1].left == -1) && (T2[R2].left == -1))//根节点都是只有右孩子，递归判断右孩子树是否同构 |
return judge(T1[R1].right, T2[R2].right); |
if (((T1[R1].left != -1) && (T2[R2].left != -1)) && ((T1[T1[R1].left].element) == (T2[T2[R2].left].element))) |
return (judge(T1[R1].left, T2[R2].left) && judge(T1[R1].right, T2[R2].right)); |
else |
return (judge(T1[R1].left, T2[R2].right) && judge(T1[R1].right, T2[R2].left)); |
} |
int main () |
{ |
Tree R1,R2; |
R1=TreeBuild(T1); |
R2=TreeBuild(T2); |
if(judge(R1,R2)) |
cout<<"Yes"<<endl; |
else |
cout<<"No"<<endl; |
} |
by: 发表于：2017-12-13 10:33:09 顶(0) | 踩(0) 回复
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