用c语言编写程序，已知一棵二叉树的先序和中序，创建这棵二叉树，并输出后序遍历的顺序

非常感谢你的问题！下面是用C语言创建一棵二叉树的代码，以及输出后序遍历的代码:

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

/* 树节点的定义 */
typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

/* 递归函数创建二叉树 */
TreeNode* buildTree(int* preorder, int prestart, int preend, int* inorder, int instart, int inend) {
    if (prestart > preend || instart > inend) {
        return NULL; // 递归结束
    }

    int root_val = preorder[prestart];
    int i;
    for (i = instart; i <= inend; i++) {
        if (inorder[i] == root_val) {
            break; // 在中序遍历中找到根节点
        }
    }

    int left_count = i - instart;
    int right_count = inend - i;
    TreeNode* root = (TreeNode*) malloc(sizeof(TreeNode));
    root->val = root_val;
    root->left = buildTree(preorder, prestart + 1, prestart + left_count, inorder, instart, i - 1);
    root->right = buildTree(preorder, preend - right_count + 1, preend, inorder, i + 1, inend);
    return root;
}

/* 后序遍历函数 */
void postorderTraversal(TreeNode* root) {
    if (root == NULL) {
        return;
    }
    postorderTraversal(root->left);
    postorderTraversal(root->right);
    printf("%d ", root->val);
}

/* 主函数 */
int main() {
    int preorder[] = {1, 2, 4, 5, 3, 6, 7};
    int inorder[] = {4, 2, 5, 1, 6, 3, 7};
    int len = sizeof(preorder) / sizeof(int);
    TreeNode* root = buildTree(preorder, 0, len - 1, inorder, 0, len - 1);
    printf("后序遍历的顺序: ");
    postorderTraversal(root);
    return 0;
}

希望我的回答对您有帮助！