给定两棵树T1和T2。如果T1可以通过若干次左右孩子互换就变成T2，则我们称两棵树是“同构”的。例如图1给出的两棵树就是同构的，因为我们把其中一棵树的结点A、B、G的左右孩子互换后，就得到另外一棵树。而图2就不是同构的。
ͼ1
ͼ2
现给定两棵树，请你判断它们是否是同构的。
输入格式:
输出格式:
如果两棵树是同构的，输出“Yes”，否则输出“No”。
输入样例1（对应图1）：
8 A 1 2 B 3 4 C 5 - D - - E 6 - G 7 - F - - H - - 8 G - 4 B 7 6 F - - A 5 1 H - - C 0 - D - - E 2 - 输出样例1:
Yes 输入样例2（对应图2）：
8 B 5 7 F - - A 0 3 C 6 - H - - D - - G 4 - E 1 - 8 D 6 - B 5 - E - - H - - C 0 2 G - 3 F - - A 1 4 输出样例2:
No #include<iostream> #include<cstdio> using namespace std; #define MAXsize 10 #define Null -1 struct TreeNode { char element; int left; int right; }T1[MAXsize], T2[MAXsize]; int Buildtree(struct TreeNode t[]) { int N, root; scanf("%d", &N); if (!N) return Null; int *check = new int[N]; for (int i = 0; i < N; i++) { check[i] = 0; } for (int i = 0; i < N; ++i) { char tcl, tcr; cin >> t[i].element >> tcl >> tcr; if (tcl != '-') { t[i].left= tcl - '0'; check[t[i].left] = 1; } else t[i].left =Null; if (tcr != '-') { t[i].right= tcr - '0'; check[t[i].right] = 1; } else t[i].right = Null; } int index; for (index = 0; index < N; ++index) { if (!check[index]) break; } root = index; return root; } int Isomorphism(int R1, int R2) { if ((R1 == Null) && (R2 == Null)) return 1; //both empty if (((R1 == Null) && (R2 != Null)) || ((R1 != Null) && (R2 == Null))) return 0;//one of them empty if (T1[R1].element != T2[R2].element) return 0; //root different if ((T1[R1].left == Null) && (T2[R2].left == Null)) return (Isomorphism(T1[R1].right, T2[R2].right)); if (((T1[R1].left != Null) && (T2[R2].left != Null)) && (T1[T1[R1].left].element == T2[T2[R2].left].element)) return (Isomorphism(T1[R1].right, T2[R2].right) && Isomorphism(T1[R1].left, T2[R2].left)); else { return (Isomorphism(T1[R1].left, T2[R2].right) && Isomorphism(T1[R1].right, T2[R2].left)); } } int main() { int R1, R2; R1 = Buildtree(T1); R2 = Buildtree(T2); if (Isomorphism(R1, R2)) printf("Yes\n"); else printf("No\n"); return 0; }