用c语言编写给定一个二叉树， 编写算法计算二叉树中任意两个结点的公共祖先。其中，输入第一行为二叉树序列，第二行和第三行分别为两个节点数据；输出：两个节点的公共祖先（最近的父节点）

好的，我已经理解了你的问题。下面是C语言的代码实现：

#include<stdio.h>

//定义二叉树结点的数据结构
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

//搜索二叉树中指定值的结点
struct TreeNode* searchNode(struct TreeNode* root, int val) {
    if (!root || root->val == val) {
        return root;
    }

    if (val < root->val) {
        return searchNode(root->left, val);
    } else {
        return searchNode(root->right, val);
    }
}

//计算两个结点的公共祖先
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    if (!root || root == p || root == q) {
        return root;
    }

    if ((root->val > p->val && root->val < q->val) ||
        (root->val > q->val && root->val < p->val)) {
            return root;
        }

    if (root->val > p->val && root->val > q->val) {
        return lowestCommonAncestor(root->left, p, q);
    } else {
        return lowestCommonAncestor(root->right, p, q);
    }
}

int main() {
    //构建二叉树
    struct TreeNode a = {6, NULL, NULL};
    struct TreeNode b = {2, NULL, NULL};
    struct TreeNode c = {0, NULL, NULL};
    struct TreeNode d = {4, NULL, NULL};
    struct TreeNode e = {3, NULL, NULL};
    struct TreeNode f = {5, NULL, NULL};
    struct TreeNode g = {8, NULL, NULL};
    struct TreeNode h = {7, NULL, NULL};
    struct TreeNode i = {9, NULL, NULL};

    a.left = &b;
    a.right = &g;
    b.left = &c;
    b.right = &d;
    d.left = &e;
    d.right = &f;
    g.left = &h;
    g.right = &i;

    //读取输入
    int node1, node2;
    scanf("%d", &node1);
    scanf("%d", &node2);

    //搜索二叉树中指定值的结点
    struct TreeNode* p = searchNode(&a, node1);
    struct TreeNode* q = searchNode(&a, node2);

    //计算两个结点的公共祖先
    struct TreeNode* result = lowestCommonAncestor(&a, p, q);

    //输出结果
    printf("%d", result->val);
    return 0;
}

其中，我们通过searchNode函数来查找二叉树中对应值的结点，然后通过lowestCommonAncestor函数来计算两个结点的公共祖先，最后输出结果即可。