Inorder predecessor and successor for a given key in BST

PROBLEM :

Inorder predecessor and successor for a given key in BST
You need to find the inorder successor and predecessor of a given key.

Example(for above tree)

Input
4  //(no of test cases)

4  // (elements whose Inorder predecessor and successor needed to be printed)
8
20
22

Output
no inorder Predecessor         //for 4
inorder Successor is 8

inorder Predecessor is 4       //for 8
inorder Successor is 10

inorder Predecessor is 14      //for 20
inorder Successor is 22

inorder Predecessor is 20      //for 22
no inorder Successor

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SIMPLE c++ IMPLEMENTATION :
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/*  Structure of a Binary Tree
struct Node
{
    int data;
    struct Node* left, *right;
}; */

/* Function to get the maximum width of a binary tree*/

void inorder_array_fill(tree* ,int [],int *) ;
int size_tree(tree* ) ;

void inorder_predecessor_successor(tree *root,tree **pre,tree **succ,int ele)
{
if(root==NULL)
return  ;

int h,i=0;
h=size_tree(root) ;

int array[h] ;

inorder_array_fill(root,array,&i) ;

for(i=0;i<h;i++)
cout<<array[i]<<" " ;

for(i=0;i<h;i++)
{
if(array[i]==ele)
{
if(i==0)
{
cout<<"\n no inorder Predecessor " ;
cout<<"\n inorder Successor is "<<array[i+1] ;
break ;
}
else if(i==h-1)
{
cout<<"\n inorder Predecessor is "<<array[i-1] ;
cout<<"\n no inorder Successor" ;
break ;
}
else
{
cout<<"\n inorder Predecessor is "<<array[i-1] ;
cout<<"\n inorder Successor is "<<array[i+1] ;
break ;
}
}
}
}

void inorder_array_fill(tree* root,int a[],int *s)
{
    if(root==NULL)
        return  ;
     
    inorder_array_fill(root->left,a,&(*s)) ;
    a[(*s)++]=root->info ;
    inorder_array_fill(root->right,a,&(*s)) ;
 
}

int size_tree(tree* root)
{
if(root==NULL)
return 0 ;

return ((size_tree(root->left))+(size_tree(root->right))+1) ;
}              

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ALTERNATE SOLUTION :
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/*  Structure of a Binary Tree
struct Node
{
    int data;
    struct Node* left, *right;
}; */

/* Function to get the maximum width of a binary tree*/

void inorder_predecessor_successor(tree *root,tree **pre,tree **succ,int ele)
{
if(root==NULL)
{
cout<<"\n No inorder predecessor and successor" ;
return ;
}
if(root->info==ele)
{
if(root->left)
{
tree* temp ;
temp=root->left ;
while(temp->right!=NULL)
temp=temp->right ;
(*pre)=temp ;
}

if(root->right)
{
tree *temp ;
temp=root->right ;
while(temp->left!=NULL)
temp=temp->left ;
(*succ)=temp ;
}
}
else
{
if(root->info>ele)
{
(*succ)=root ;
inorder_predecessor_successor(root->left,pre,succ,ele) ;
}
else
{
(*pre)=root ;
inorder_predecessor_successor(root->right,pre,succ,ele) ;
}
}
}                


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