Construct Tree from given Inorder and Preorder Traversals

March 25, 2023

Construct Tree From Given Inorder and Preorder Traversals in Java

There are three types of traversals in a tree: Inorder, Preorder and Postorder traversal. A tree can be formed with any two tree traversals in which one of them being the in order traversal.

Given Inorder and Preorder Tree Traversals, construct the Tree.

Example

Given traversals:
Preorder: 50, 25, 12, 37, 30, 40, 75, 62, 60, 70, 87
Inorder:12, 25, 30, 37, 40, 50, 60, 62, 70, 75, 87
1. The last element in the preorder traversal is the root of the tree.
So, here 50 will be the root of the tree.
2. We will find the index of 50 in the inorder traversal.The index found is 5. Let this index is denoted by ‘pos’.
3. All the elements to the left of this index ( from 0 to 4 index) will be in the left subtree of 50.
4. And all the elements to the right of this index ( from 6 to 10) will be in the right subtree of 50.

Find size of the Binary tree

Find the height of binary tree

Find maximum in binary tree

Check whether two tree are identical

Spiral Order traversal of Tree

Now, we will divide preorder and inorder array in two parts.
One is for the left subtree and other is for the right subtree.

Let

psi: starting index for preorder array
pei: ending index for preorder array
isi: starting index of inorder array
iei: ending index of preorder array
clc: Number of elements in the left subtree

Clearly, clc= pos – isi;

For left subtree:
Preorder array: from index psi + 1, psi + clc
Inorder array: from index isi, isi + clc -1

For right subtree:
Preorder array: from index psi + clc + 1, pei
Inorder array: from index isi + clc + 1, iei

Using the above arrays, all the steps are recursively repeated.

The following binary tree is obtained:

Code For Binary Search Tree

Run

public class Main
{

// Binary tree class
  public static class BinaryTree
  {
// Node class
    public class Node
    {
      int data;
      Node left;
      Node right;

      public Node (int data)
      {
	this.data = data;
	this.left = null;
	this.right = null;
      }
    }

    private Node root;
    private int size;

    public BinaryTree (int[]pre, int[]in)
    {
      this.root =
	this.construct (pre, 0, pre.length-1, in, 0, in.length-1);
    }

    private Node construct (int[]pre, int psi, int pei, int[]in, int isi,int iei)
    {

//Base case
      if (psi > pei)
	{
	  return null;
	}
      Node node = new Node (pre[psi]);
      node.left = null;
      node.right = null;
      this.size++;

//Finding pre[psi] in the inorder array
      int pos = -1;
      for (int i = isi; i <= iei; i++)
	{
	  if (in[i] == node.data)
	    {
	      pos = i;
	      break;
	    }
	}

//Number of elements in left subtree
      int clc = pos-isi;

//Left subtree
      node.left =
	this.construct (pre, psi + 1, psi + clc, in, isi, isi + clc-1);

//Right subtree
      node.right = this.construct (pre, psi + clc + 1, pei, in, isi + clc + 1,
				   iei);

      return node;
    }

// Preorder tree traversal
    public void preOrder ()
    {
      preOrder (this.root);
    }

    private void preOrder (Node node)
    {
      if (node == null)
	{
	  return;
	}

      System.out.print (node.data + " ");
      preOrder (node.left);
      preOrder (node.right);

    }
  }

  public static void main (String[]args) throws Exception
  {

// Construct binary tree

    int[] in = { 12, 25, 30, 37, 40, 50, 60, 62, 70, 75, 87 };
    int[] pre = { 50, 25, 12, 37, 30, 40, 75, 62, 60, 70, 87 };

    BinaryTree bt = new BinaryTree (pre, in);
    System.out.println("The new tree constructed is :");
    bt.preOrder ();

  }
}

Output :

The new tree constructed is :
50 25 12 37 30 40 75 62 60 70 87

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