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 <stdio.h>
#include <stdlib.h>
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
递归这种思想,很有14
写递归的技巧是管好当下,之后的事抛给递归。具体到这里,不管 N 是多少,当前的选择只有两个:匹配 0 次、匹配 1 次。所以可以这样处理:
空根刚刚好,肯定是对的。
一空一不空,&&或者俩元素不一样肯定不对呀
左左 右右配对||,或者 左右 右左配对上一个就行了
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));
}