Construct Special Binary Tree from given Inorder traversal

PROBLEM :

Given Inorder Traversal of a Special Binary Tree in which key of every node is greater than keys in left and right children, construct the Binary Tree and return root.

Examples:

Input: inorder[] = {5, 10, 40, 30, 28}
Output: root of following tree
         40
       /   \
      10     30
     /         \
    5          28

Input: inorder[] = {1, 5, 10, 40, 30,
                    15, 28, 20}
Output: root of following tree
          40
        /   \
       10     30
      /         \
     5          28
    /          /  \
   1         15    20

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SIMPLE c++ IMPLEMENTATION :
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/*
typedef struct BST
{
int info ;
struct BST *left ;
struct BST *right ;
}tree ;------------------------------------*/

tree * construct_special_tree(int [],int ,int) ;
int find_max_arrayPos(int [],int ,int ) ;

tree * special_Btree()
{
int no ;
cout<<"\n enter the number of element in the inorder array" ;
cin>>no ;

int a[no],i ;
for(i=0;i<no;i++)
cin>>a[i] ;

tree *root ;
root=construct_special_tree(a,0,no-1) ;
return root  ;
}

tree * construct_special_tree(int a[],int start,int end)
{
if(start>end)
return NULL ;

tree *temp=(tree*)malloc(sizeof(tree)) ;
temp->left=NULL ;
temp->right=NULL ;

if(start==end)
{
temp->info=a[start] ;
return temp ;
}

int p ;
p=find_max_arrayPos(a,start,end) ;

temp->info=a[p] ;
temp->left=construct_special_tree(a,start,p-1) ;
temp->right=construct_special_tree(a,p+1,end) ;

return temp ;
}

int find_max_arrayPos(int a[],int start,int end)
{
int i ,maxi;
maxi=start ;
for(i=start;i<=end;i++)
{
if(a[i]>a[maxi])
maxi=i ;
}
return maxi ;
}

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