c语言实现判断两颗树是否同构

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在本题中认为如果两个树左右子树交换可以相同，也被认为是同构树。

对应输入格式为：4(总结点数)

A - 1

B 2 3

C - -

D - -

#include <stdio.h>#define Tree int#define Null -1#define MAXSIZE 10struct Node{char Element;Tree Left;Tree Right;}T1[MAXSIZE], T2[MAXSIZE];Tree BuildTree(struct Node T[]){int N, i, Root;char ch, cl, cr;scanf("%d", &N);ch = getchar();int Check[N];for(i = 0; i < N; i++) Check[i] = 0;//为了标记根节点 ，因为没有节点指向根节点 for(i = 0; i < N; i++){scanf("%c %c %c", &T[i].Element, &cl, &cr);ch = getchar();if(cl != '-'){T[i].Left = cl - '0';Check[T[i].Left] = 1;}elseT[i].Left = Null;if(cr != '-'){T[i].Right = cr - '0';Check[T[i].Right] = 1;}elseT[i].Right = Null;}for(i = 0; i < N; i++)//遍历找到根节点 if(Check[i] == 0)break;Root = i;return Root;}int Isomorphic(Tree R1, Tree R2){//两树都为空 if(R1 == Null && R2 == Null)return 1;//空树和非空树if((R1 == Null && R2 != Null) || (R1 != Null && R2 == Null))return 0;//两树的根节点不一样if(T1[R1].Element != T2[R2].Element)return 0;//若过两树的左子树都空，判断右子树是否一样if(T1[R1].Left == Null && T2[R2].Left == Null)return Isomorphic(T1[R1].Right, T2[R2].Right);//若两树的左子树不空，并且左子树的结点元素都一样，判断左右子树是否一样if((T1[R1].Left != Null && T2[R2].Left != Null) && (T1[T1[R1].Left].Element == T2[T2[R2].Left].Element))return Isomorphic(T1[R1].Left, T2[R2].Left) && Isomorphic(T1[R1].Right, T2[R2].Right);else//否则 判断两树是否同构 return Isomorphic(T1[R1].Left, T2[R2].Right) && Isomorphic(T1[R1].Right, T2[R2].Left);}int main(){Tree R1, R2;R1 = BuildTree(T1);R2 = BuildTree(T2);if(Isomorphic(R1, R2)) printf("Yes");elseprintf("No");return 0;}

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