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 = -;
         for(int i = istart; i <= iend; i ++) {
             if(inorder[i] == preorder[pstart]){
                 idx_root = i;
                 break;
             }
         }

         if(idx_root == -)
             return NULL;

         int left_size = idx_root - istart;
         if(left_size >  )
             root->left = build(preorder, pstart + , pstart + left_size, inorder, istart, idx_root - );

         int right_size = iend - idx_root;
         if(right_size > )
             root->right = build(preorder, pend - right_size +  , pend, inorder, idx_root + , iend);

         return root;
     }

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

         return build(preorder, , preorder.size() -, inorder, , inorder.size() - );
     }