Pre-order to post-order traversal

Question

If the pre-order traversal of a binary search tree is 6, 2, 1, 4, 3, 7, 10, 9, 11, how to get the post-order traversal?

Answer 1

Based on Ondrej Tucny's answer. Valid for BST only
example:

     20  
    /  \  
   10  30  
   /\    \  
  6  15   35  

Preorder = 20 10 6 15 30 35
Post = 6 15 10 35 30 20

For a BST, In Preorder traversal; first element of array is 20. This is the root of our tree. All numbers in array which are lesser than 20 form its left subtree and greater numbers form right subtree.

//N = number of nodes in BST (size of traversal array)
int post[N] = {0}; 
int i =0;

void PretoPost(int pre[],int l,int r){
  if(l==r){post[i++] = pre[l]; return;}
  //pre[l] is root
  //Divide array in lesser numbers and greater numbers and then call this function on them recursively  
  for(int j=l+1;j<=r;j++) 
      if(pre[j]>pre[l])
          break;
  PretoPost(a,l+1,j-1); // add left node
  PretoPost(a,j,r); //add right node
  //root should go in the end
  post[i++] = pre[l]; 
  return;
 }

Please correct me if there is any mistake.