4-11 Isomorphic

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.

如果T1可以通过交换T1中（一些）节点的左右子节点而转换为T2，那么T1和T2这两棵树是同构的。例如，图1中的两棵树就是同构的，因为如果交换了A、B和G的子节点而不是其他节点的子节点，它们是相同的。给出一个多项式时间算法来判断两棵树是否同构。

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.

如果T1和T2确实同构，则函数应该返回1，否则返回0。

Sample program of judge:

#include 
#include 
 
typedef char ElementType;
 
typedef struct TreeNode *Tree;
struct TreeNode {
    ElementType Element;
    Tree  Left;
    Tree  Right;
};
 
Tree BuildTree(); /* details omitted */
 
int Isomorphic( Tree T1, Tree T2 );
 
int main()
{
    Tree T1, T2;
    T1 = BuildTree();
    T2 = BuildTree();
    printf(“%d\n”, Isomorphic(T1, T2));
    return 0;
}
 
/* Your function will be put here */

Sample Output 1 (for the trees shown in Figure 1):

1

Sample Output 2 (for the trees shown in Figure 2):

0

---

代码：

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));
}

递归方法：

如果两个树都是空树的时候，那么同构

if (T1 == NULL && T2 == NULL) 
        return 1;

因为上一个if语句判断了左右孩子都是空的情况，这里如果一个根节点是空，另外一个不为空，或者两个根节点的值不同，那么不是同构的

else if (T1 == NULL || T2 == NULL || T1->Element != T2->Element) 
        return 0;

同构的情况有两种，一种是交换了：

(Isomorphic(T1->Right, T2->Left) && Isomorphic(T1->Left, T2->Right))

一种是没交换的：

Isomorphic(T1->Left, T2->Left) && Isomorphic(T1->Right, T2->Right)

这两种情况 出现一种就行：

return 
        Isomorphic(T1->Left, T2->Left) && Isomorphic(T1->Right, T2->Right) || (Isomorphic(T1->Right, T2->Left) && Isomorphic(T1->Left, T2->Right));