Construct Binary Tree from Preorder and Inorder Traversal [LeetCode]

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

Solution: Just do it recursively.

    TreeNode *build(vector<int> &preorder, int pstart, int pend, vector<int> &inorder, int istart, int iend) {
        TreeNode * root = new TreeNode(preorder[pstart]); 
        int idx_root = -1;
        for(int i = istart; i <= iend; i ++) {
            if(inorder[i] == preorder[pstart]){
                idx_root = i;
                break;
            }
        }
        
        if(idx_root == -1)
            return NULL;
            
        int left_size = idx_root - istart;
        if(left_size > 0 )
            root->left = build(preorder, pstart + 1, pstart + left_size, inorder, istart, idx_root - 1);
            
        int right_size = iend - idx_root;
        if(right_size > 0)
            root->right = build(preorder, pend - right_size + 1 , pend, inorder, idx_root + 1, iend);
        
        return root;
    }

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size() == 0 || inorder.size() == 0 || preorder.size() != inorder.size())
            return NULL;
        
        return build(preorder, 0, preorder.size() -1, inorder, 0, inorder.size() - 1);
    }