C语言求任意二叉树中第一条最长的路径长度，并输出此路径上各结点的值。

这道题可以用递归的方式来解决，具体思路如下：

1. 如果二叉树为空，直接返回0。

2. 分别计算左右子树的最大深度maxDepth和左右子树中的最长路径长度，分别为leftPath和rightPath。

3. 比较leftPath + rightPath和当前最长路径长度maxLength，如果大于maxLength，则更新maxLength，并记录当前路径上的结点值。

4. 最终返回maxDepth和maxLength中的较大值作为当前子树的最大深度。

下面是对应的C语言代码实现：

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

// 二叉树结点定义
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

void findLongestPath(struct TreeNode* root, int* maxLength, int** path, int* pathIndex) {
    if (!root) {
        return;
    }
  
    int leftDepth = findLongestPath(root->left, maxLength, path, pathIndex);
    int rightDepth = findLongestPath(root->right, maxLength, path, pathIndex);

    int leftPath = 0;
    if (root->left && root->left->val == root->val) {
        leftPath = leftDepth + 1;
    }

    int rightPath = 0;
    if (root->right && root->right->val == root->val) {
        rightPath = rightDepth + 1;
    }

    int currentPath = leftPath + rightPath;
    if (currentPath > *maxLength) {
        *maxLength = currentPath;
        *pathIndex = 0;
        (*path)[(*pathIndex)++] = root->val;
    } else if (currentPath == *maxLength) {
        (*path)[(*pathIndex)++] = root->val;
    }
  
    return 1 + (leftDepth > rightDepth ? leftDepth : rightDepth);
}

int* longestUnivaluePath(struct TreeNode* root, int* returnSize) {
    int maxLength = 0;
    int* path = (int*)malloc(sizeof(int) * 1000);
    int pathIndex = 0;

    findLongestPath(root, &maxLength, &path, &pathIndex);

    *returnSize = pathIndex;
    return path;
}

int main() {
    // 构造一棵二叉树
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = 1;
    root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->left->val = 4;
    root->left->left = NULL;
    root->left->right = NULL;
    root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->val = 5;
    root->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->left->val = 5;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->val = 5;
    root->right->right->left = NULL;
    root->right->right->right = NULL;

    int returnSize;
    int* path = longestUnivaluePath(root, &returnSize);

    printf("The longest path length is: %d\n", returnSize);
    printf("The values on the path are: ");
    for (int i = 0; i < returnSize; i++) {
        printf("%d ", path[i]);
    }
    printf("\n");

    free(path);
    free(root->left);
    free(root->right->left);
    free(root->right->right);
    free(root);
    return 0;
}

注：此代码中的二叉树是固定的，实际使用时需要根据题目要求构造对应的二叉树。