How to infer a preposition traversal in a tree, taking into account the transformation of order and after order?

Question

Given the code for outputting a tree post-order traversal when I have pre-order and order traversal in an integer array. How do I also get pre-order with given inorder and postorder arrays?

void postorder( int preorder[], int prestart, int inorder[], int inostart, int length)
{ 
  if(length==0) return; //terminating condition
  int i;
  for(i=inostart; i<inostart+length; i++)
    if(preorder[prestart]==inorder[i])//break when found root in inorder array
      break;
  postorder(preorder, prestart+1, inorder, inostart, i-inostart);
  postorder(preorder, prestart+i-inostart+1, inorder, i+1, length-i+inostart-1);
  cout<<preorder[prestart]<<" ";
}

Here is a prototype of preorder ()

void preorder (int inorderorder [], int inostart, int postorder [], int poststart, int length)

use postorder () will be

int preorder[6]={6,4,1,5,8,9};
int inorder[6]={1,4,5,6,8,9};
postorder( preorder,0,inorder,0,6);

out put will

1 5 4 9 8 6

below is the wrong code for print_preorder () which still doesn't work below

void print_preorder( int inorder[], int inostart, int postorder[], int poststart, int length)
    {
      if(length==0) return; //terminating condition
      int i;
      for(i=inostart; i<inostart+length; i++)
        if(postorder[poststart+length-1]==inorder[i])
          break; 
      cout<<postorder[poststart+length-1]<<" ";
      print_preorder(inorder, inostart , postorder, poststart, i-inostart);
      print_preorder(inorder, inostart+i-poststart+1, postorder, i+1, length-i+inostart-1);
    }

Answer 1

Here are some suggestions:

• postorder preorder.
• inorder preorder.
• print_preorder inorder.
• print_preorder inorder postorder .
• : postorder[poststart+length] . , postorder[poststart+length-1]
• print_preorder . , length - , inostart - inorder. , length.
• . , .

:

     6
   /   \
  4     8
 / \     \
1   5     9

:

// index:         0 1 2 3 4 5
int postorder[6]={1,5,4,9,8,6};
int inorder[6]=  {1,4,5,6,8,9};
int preorder[6]= {6,4,1,5,8,9};

, , :)

( ):

6 print_preorder(len=6, in=[1 4 5 6 8 9], post=[1 5 4 9 8 6])
4 |-> print_preorder(len=3, in=[1 4 5], post=[1 5 4])
1 |   |-> print_preorder(len=1, in=[1], post=[1])
  |   |   |-> print_preorder(len=0, in=[], post=[])
  |   |   |-> print_preorder(len=0, in=[], post=[])
5 |   |-> print_preorder(len=1, in=[5], post=[5])
  |       |-> print_preorder(len=0, in=[], post=[])
  |       |-> print_preorder(len=0, in=[], post=[])
8 |-> print_preorder(len=2, in=[8 9], post=[9 8])
      |-> print_preorder(len=0, in=[], post=[])
9     |-> print_preorder(len=1, in=[9], post=[9])
          |-> print_preorder(len=0, in=[], post=[])
          |-> print_preorder(len=0, in=[], post=[])

:)

Answer 2

• post order .

• Inorder, . - , - .

• , , .

• . inorder. , .

  public static void printpreorder(char []inorder,int i_start,int i_end,char[] postorder,int p_start,int p_end)
  {
    if(i_start > i_end || p_start > p_end)
           return ; 
    char root = postorder[p_end];
    System.out.print(root);
    System.out.print("(");
      int k=0;
        for(int i=0; i< inorder.length; i++){
            if(inorder[i]==root){
              k = i;
              break;
            }
        }
    printpreorder(inorder, i_start, k-1, postorder, p_start, p_start+k-(i_start+1));
    System.out.print(")(");
    printpreorder(inorder, k+1, i_end, postorder, p_start+k-i_start, p_end-1);
    System.out.print(")");    
  }
  

. @Stephen