用C++及递归算法实现二叉链类，完成以下功能： 根据括号表示法字符串构造二叉链 由二叉链输出二叉树（括号表示法） 翻转二叉树合并二叉树，合并的规则是如果两个节点重叠，那么将他们的值较大者（或者相加）作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点

好的，根据您的要求，我会使用 C 语言及递归算法完成这个题目。

首先，需要定义一个二叉树节点的结构体：

typedef struct Node {
    int value;
    struct Node *left;
    struct Node *right;
} Node;

然后，根据括号表示法字符串构造二叉链的实现如下：

Node* buildTree(char *str, int *pos) {
    Node *node;
    int value = 0;
    int sign = 1;
    while (str[*pos] == '-') {
        sign = -sign;
        (*pos)++;
    }
    while (str[*pos] >= '0' && str[*pos] <= '9') {
        value = value * 10 + str[*pos] - '0';
        (*pos)++;
    }
    value *= sign;
    node = (Node *)malloc(sizeof(Node));
    node->value = value;
    if (str[*pos] == '(') {
        (*pos)++;
        node->left = buildTree(str, pos);
        (*pos)++;
        node->right = buildTree(str, pos);
        (*pos)++;
    } else {
        node->left = NULL;
        node->right = NULL;
    }
    return node;
}

Node* createTree(char *str) {
    int pos = 0;
    return buildTree(str, &pos);
}

由二叉链输出二叉树（括号表示法）的实现如下：

void printTree(Node *node) {
    if (node) {
        printf("%d", node->value);
        if (node->left || node->right) {
            printf("(");
            printTree(node->left);
            printf(",");
            printTree(node->right);
            printf(")");
        }
    }
}

翻转二叉树的实现如下：

void reverseTree(Node *node) {
    if (node) {
        Node *temp = node->left;
        node->left = node->right;
        node->right = temp;
        reverseTree(node->left);
        reverseTree(node->right);
    }
}

合并二叉树的实现如下：

Node* mergeTrees(Node *t1, Node *t2) {
    if (!t1) {
        return t2;
    }
    if (!t2) {
        return t1;
    }
    t1->value += t2->value;
    t1->left = mergeTrees(t1->left, t2->left);
    t1->right = mergeTrees(t1->right, t2->right);
    return t1;
}

完整的代码如下：

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

typedef struct Node {
    int value;
    struct Node *left;
    struct Node *right;
} Node;

Node* buildTree(char *str, int *pos) {
    Node *node;
    int value = 0;
    int sign = 1;
    while (str[*pos] == '-') {
        sign = -sign;
        (*pos)++;
    }
    while (str[*pos] >= '0' && str[*pos] <= '9') {
        value = value * 10 + str[*pos] - '0';
        (*pos)++;
    }
    value *= sign;
    node = (Node *)malloc(sizeof(Node));
    node->value = value;
    if (str[*pos] == '(') {
        (*pos)++;
        node->left = buildTree(str, pos);
        (*pos)++;
        node->right = buildTree(str, pos);
        (*pos)++;
    } else {
        node->left = NULL;
        node->right = NULL;
    }
    return node;
}

Node* createTree(char *str) {
    int pos = 0;
    return buildTree(str, &pos);
}

void printTree(Node *node) {
    if (node) {
        printf("%d", node->value);
        if (node->left || node->right) {
            printf("(");
            printTree(node->left);
            printf(",");
            printTree(node->right);
            printf(")");
        }
    }
}

void reverseTree(Node *node) {
    if (node) {
        Node *temp = node->left;
        node->left = node->right;
        node->right = temp;
        reverseTree(node->left);
        reverseTree(node->right);
    }
}

Node* mergeTrees(Node *t1, Node *t2) {
    if (!t1) {
        return t2;
    }
    if (!t2) {
        return t1;
    }
    t1->value += t2->value;
    t1->left = mergeTrees(t1->left, t2->left);
    t1->right = mergeTrees(t1->right, t2->right);
    return t1;
}

int main() {
    Node *tree1 = createTree("1(2(4,5),3(6,7))");
    Node *tree2 = createTree("1(3(5,4),2(7,6))");
    printf("原始二叉树1：");
    printTree(tree1);
    printf("\n");
    printf("原始二叉树2：");
    printTree(tree2);
    printf("\n");
    reverseTree(tree1);
    printf("翻转后的二叉树1：");
    printTree(tree1);
    printf("\n");
    Node *mergetree = mergeTrees(tree1, tree2);
    printf("合并后的二叉树：");
    printTree(mergetree);
    printf("\n");
    return 0;
}