力扣106：从中序和后序序列构造二叉树

给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

示例 2:

输入：inorder = [-1], postorder = [-1]
输出：[-1]

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize) {
    if (postorderSize <= 0 || inorderSize <= 0) return NULL;
    
   
    struct TreeNode *root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->val = postorder[postorderSize - 1];
    

    int index;
    for (index = 0; index < inorderSize; index++) {
        if (inorder[index] == postorder[postorderSize - 1]) {
            break;
        }
    }
    

    root->left = buildTree(inorder, index, postorder, index);
    root->right = buildTree(inorder + index + 1, inorderSize - index - 1, postorder + index, postorderSize - 1 - index);
    
    return root;
}