6-7 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.

                                           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(); /* 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

                                    Figure2

 

具体代码实现为：

int Isomorphic(Tree T1,Tree T2)
{
 	if((T1 == NULL) && (T2 == NULL))  //树都为空树，视为同构 
 	    return 1;
 	if(((T1 == NULL)&&(T2 != NULL)) || ((T1 != NULL)&&(T2 == NULL)))  //两树其中之一为空树,不同构 
	    return 0;
	if(T1->Element != T2->Element)  //若两树都非空的话，比较根结点 
	    return 0;
	if((T1->Left == NULL) && (T2->Left == NULL))  //左子树都为空，比较右子树 
	    return Isomorphic(T1->Right,T2->Right);
    if((T1->Left != NULL)&&(T2->Left != NULL) && (T1->Left->Element == T2->Left->Element))
    /*左子树非空，并且左子树根结点相同，递归比较左右子树*/ 
        return (Isomorphic(T1->Left,T2->Left) && Isomorphic(T1->Right,T2->Right));
    else
        return (Isomorphic(T1->Left,T2->Right) && Isomorphic(T1->Right,T2->Left));
 }