4-11 Isomorphic (10 分)

4-11 Isomorphic (10 分)
Two trees, T1 and T2, are isomorphic if T1 can be transformed into T2 by swapping left and right children of (some of the) nodes in T1. For instance, the two trees in Figure 1 are isomorphic because they are the same if the children of A, B, and G, but not the other nodes, are swapped. Give a polynomial time algorithm to decide if two trees are isomorphic.

Figure 1
Format of functions:
int Isomorphic( Tree T1, Tree T2 );
where Tree is defined as the following:

typedef struct TreeNode *Tree;
struct TreeNode {
ElementType Element;
Tree Left;
Tree Right;
};
The function is supposed to return 1 if T1 and T2 are indeed isomorphic, or 0 if not.

Sample program of judge:
#include
#include

typedef char ElementType;

typedef struct TreeNode *Tree;
struct TreeNode {
ElementType Element;
Tree Left;
Tree Right;
};

Tree BuildTree();

int Isomorphic( Tree T1, Tree T2 );

int main()
{
Tree T1, T2;
T1 = BuildTree();
T2 = BuildTree();
printf(“%dn”, Isomorphic(T1, T2));
return 0;
}

Sample Output 1 (for the trees shown in Figure 1):
1
结尾无空行
Sample Output 2 (for the trees shown in Figure 2):
0

Figure2

先说一下定义两棵树同构就是通过交换左右子树，两棵树可以交换到相等的形式。
算法就是先建树，再判断是否同构
同构可以通过递归实现
分为三种情况
（1）空根直接返回1

（2）一空一不空或者或者两个根元素不一样返回0

（3）不是上面两种情况就可以递归到下一层进行了，左左 右右配对或者 左右 右左配对，有一个成立就可以了

int Isomorphic( Tree T1, Tree T2 ){
    if(T1==NULL&&T2==NULL) return 1;
    else if(T1==NULL||T2==NULL||T1->Element!=T2->Element) return 0;
    return Isomorphic(T1->Left,T2->Left)&&Isomorphic(T1->Right,T2->Right)||(Isomorphic(T1->Right,T2->Left)&&Isomorphic(T1->Left,T2->Right));
}