假设一棵二叉树的先序序列为EBADCFHGIKJ，中序序列为ABCDEFGHIJK，请画出该二叉树，并写出后序序列。C语言代码

二叉树的节点结构体定义如下：

struct TreeNode {
    char val;
    struct TreeNode* left;
    struct TreeNode* right;
};

可以通过递归构建二叉树，具体实现如下：

struct TreeNode* buildTree(char* preorder, int preorderSize, char* inorder, int inorderSize) {
    return build(preorder, 0, preorderSize - 1, inorder, 0, inorderSize - 1);
}

struct TreeNode* build(char* preorder, int preStart, int preEnd, char* inorder, int inStart, int inEnd) {
    if (preStart > preEnd || inStart > inEnd) {
        return NULL;
    }
    char rootVal = preorder[preStart];
    int index = 0;
    for (int i = inStart; i <= inEnd; i++) {
        if (inorder[i] == rootVal) {
            index = i;
            break;
        }
    }
    int leftSize = index - inStart;
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = rootVal;
    root->left = build(preorder, preStart + 1, preStart + leftSize, inorder, inStart, index - 1);
    root->right = build(preorder, preStart + leftSize + 1, preEnd, inorder, index + 1, inEnd);
    return root;
}

构建完成后，可以进行后序遍历，得到后序序列：

void postorder(struct TreeNode* root, char* result, int* returnSize) {
    if (root == NULL) {
        return;
    }
    postorder(root->left, result, returnSize);
    postorder(root->right, result, returnSize);
    result[(*returnSize)++] = root->val;
}

char* postorderTraversal(struct TreeNode* root, int* returnSize) {
    char* result = (char*)malloc(sizeof(char) * 1000);
    *returnSize = 0;
    postorder(root, result, returnSize);
    return result;
}