用C语言写一段判断二叉树左右子树是否同构的代码

好的，以下是用C语言写的判断二叉树左右子树是否同构的代码：

#include <stdio.h>
#include <stdlib.h>

typedef struct node {
    char val;
    struct node* left;
    struct node* right;
} Node;

int isIsomorphic(Node* root1, Node* root2) {
    // 如果两个节点都为空，则认为它们同构
    if (!root1 && !root2) {
        return 1;
    }
    // 如果只有一个节点为空，则认为它们不同构
    if (!root1 || !root2) {
        return 0;
    }
    // 如果节点的值不同，则认为它们不同构
    if (root1->val != root2->val) {
        return 0;
    }
    // 分别比较左右子树
    return (isIsomorphic(root1->left, root2->left) && isIsomorphic(root1->right, root2->right)) ||
           (isIsomorphic(root1->left, root2->right) && isIsomorphic(root1->right, root2->left));
}

int main() {
    // 构造两棵二叉树
    Node* root1 = (Node*)malloc(sizeof(Node));
    Node* root2 = (Node*)malloc(sizeof(Node));
    root1->val = 'a';
    root1->left = (Node*)malloc(sizeof(Node));
    root1->left->val = 'b';
    root1->left->left = NULL;
    root1->left->right = (Node*)malloc(sizeof(Node));
    root1->left->right->val = 'c';
    root1->left->right->left = NULL;
    root1->left->right->right = NULL;
    root1->right = (Node*)malloc(sizeof(Node));
    root1->right->val = 'd';
    root1->right->left = (Node*)malloc(sizeof(Node));
    root1->right->left->val = 'e';
    root1->right->left->left = NULL;
    root1->right->left->right = NULL;
    root1->right->right = (Node*)malloc(sizeof(Node));
    root1->right->right->val = 'f';
    root1->right->right->left = NULL;
    root1->right->right->right = NULL;

    root2->val = 'a';
    root2->left = (Node*)malloc(sizeof(Node));
    root2->left->val = 'd';
    root2->left->left = (Node*)malloc(sizeof(Node));
    root2->left->left->val = 'f';
    root2->left->left->left = NULL;
    root2->left->left->right = NULL;
    root2->left->right = (Node*)malloc(sizeof(Node));
    root2->left->right->val = 'e';
    root2->left->right->left = NULL;
    root2->left->right->right = NULL;
    root2->right = (Node*)malloc(sizeof(Node));
    root2->right->val = 'b';
    root2->right->left = NULL;
    root2->right->right = (Node*)malloc(sizeof(Node));
    root2->right->right->val = 'c';
    root2->right->right->left = NULL;
    root2->right->right->right = NULL;

    // 判断两棵二叉树是否同构
    if (isIsomorphic(root1, root2)) {
        printf("两个二叉树同构\n");
    } else {
        printf("两个二叉树不同构\n");
    }

    // 释放内存
    free(root1->left->right);
    free(root1->left);
    free(root1->right->left);
    free(root1->right->right);
    free(root1->right);
    free(root1);

    free(root2->left->left);
    free(root2->left->right);
    free(root2->left);
    free(root2->right->right);
    free(root2->right);
    free(root2);

    return 0;
}