PU Construct Binary Tree from Preorder and Inorder Traversal

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize) {
    if (!preorderSize) return 0;
    struct TreeNode *root = malloc(sizeof(struct TreeNode));
    root->val = preorder[0];
    int i;
    for (i = 0; i < inorderSize && inorder[i] != preorder[0]; i++);
    root->left = buildTree(preorder + 1, i, inorder, i);
    root->right = buildTree(preorder + 1 + i, preorderSize - 1 - i, inorder + i + 1, inorderSize - i - 1);
    return root;
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize) {
    if (preorderSize < 1) return 0;
    struct TreeNode *root = malloc(sizeof(struct TreeNode)), *cur = root;
    root->val = preorder[0];
    root->left = 0;
    root->right = 0;
    struct TreeNode **stack = malloc(preorderSize * sizeof(struct TreeNode *));
    stack[0] = root;
    int top = 1, i, j = 0, isright = 0;
    for (i = 1; i < preorderSize; i++) {
        while (top && stack[top - 1]->val == inorder[j]) {
            cur = stack[--top];
            isright = 1;
            j++;
        }
        struct TreeNode *child = malloc(sizeof(struct TreeNode));
        child->val = preorder[i];
        child->left = 0;
        child->right = 0;
        stack[top++] = child;
        if (isright) {
            cur->right = child;
            isright = 0;
        }
        else {
            cur->left = child;
        }
        cur = child;
    }
    return root;
}