Construct BST from given preorder traversal | Set 2 | GeeksforGeeks

Construct BST from given preorder traversal | Set 2 | GeeksforGeeks
Related: Construct a Binary Search Tree from given postorder
Given preorder traversal of a binary search tree, construct the BST.

1. Create an empty stack.

2. Make the first value as root. Push it to the stack.

3. Keep on popping while the stack is not empty and the next value is greater than stack’s top value. Make this value as the right child of the last popped node. Push the new node to the stack.

4. If the next value is less than the stack’s top value, make this value as the left child of the stack’s top node. Push the new node to the stack.

Time Complexity: O(n). The complexity looks more from first look. If we take a closer look, we can observe that every item is pushed and popped only once. So at most 2n push/pop operations are performed in the main loops of constructTree(). Therefore, time complexity is O(n).
http://algorithms.tutorialhorizon.com/construct-binary-search-tree-from-a-given-preorder-traversal-using-recursion/

Solu­tion to the prob­lem is sim­i­lar to isBST Max-Min Solu­tion.

“Your root value can have any value between -∞ to + ∞, say it is 30 here, When you val­i­date the right child of 30, it can take any value between 30 and + ∞. When you val­i­date the left child of 30, it can take any value between — ∞ and 30. likewise

So the idea is Pass the min­i­mum and max­i­mum val­ues between which the node’s value must lie.

• Exam­ple: int[] pre­Order = { 20, 10, 5, 1, 7, 15, 30, 25, 35, 32, 40 };
• First ele­ment in the pre­order[] will def­i­nitely be the root, which is 20 here.
• we start with the range min­i­mum = Integer.MIN_VALUE and max­i­mum = Interger.MAX_VALUE, so your root can take any value between this range.
• So when putting the left node of 20(root), it must lie within the range to min­i­mum = Integer.MIN_VALUE and max­i­mum = 20, so check the next ele­ment in the pre­order[], if it lies in this range, make it the left child to the root, else it must the the right chlid of the root and so on. See the fig­ure for bet­ter under­stand­ing. ( see the exe­cu­tion sequence)
• Solve it recursively.

public int pIndex = 0;

public Node constructTree(int[] preorder, int data, int min, int max) {
  if (pIndex < preorder.length) {
    if (preorder[pIndex] > min && preorder[pIndex] < max) {
      Node root = new Node(data);
      pIndex++;
      if (pIndex < preorder.length) { // if not needed
        // nodes lies between min and data will create left subtree
        root.left = constructTree(preorder, preorder[pIndex], min,
            data);
        // nodes lies between data and max will create right subtree
        root.right = constructTree(preorder, preorder[pIndex],
            data, max);
      }
      return root;
    }
  }
  return null;
}
Node root = p.constructTree(preOrder, preOrder[0], Integer.MIN_VALUE,
    Integer.MAX_VALUE);

http://algorithms.tutorialhorizon.com/construct-binary-search-tree-from-a-given-preorder-traversal-using-stack-without-recursion/

• Exam­ple: int[] pre­Order = { 20, 10, 5, 1, 7, 15, 30, 25, 35, 32, 40 };
• Use Stack.
• First ele­ment in the pre­order[] will def­i­nitely be the root, which is 20 here.
• Cre­ate a node with data 20 and push it in Stack.
• Now take the next ele­ment (say ‘data’) from the pre­order sequence.
• If ‘data’ is greater than the top item in the stack then keep pop­ping out the nodes from the stack, keep stor­ing it in temp node till the top node in stack is less than the ‘data’.
• cre­ate a node with ‘data’ and add it to the right of temp node and push the temp node to stack.
• If ‘data’ is less than the top item in the stack then cre­ate a node with ‘data’ and add it to the left of top node in stack and push it in the stack.
• Repeat the above two steps till all the ele­ments in pre­order[] is over.
• return the root

public Node constructTree(int[] preorder) {
  Stack<Node> s = new Stack<Node>();
  Node root = new Node(preorder[0]);
  s.add(root);
  for (int i = 1; i < preorder.length; i++) {
    Node x = null;
    while (!s.isEmpty() && preorder[i] > s.peek().data) {
      x = s.pop();
    }
    if (x != null) {
      x.right = new Node(preorder[i]);
      s.push(x.right);
    } else {
      s.peek().left = new Node(preorder[i]);
      s.push(s.peek().left);
    }
  }
  return root;
}

Node* constructTree ( int pre[], int size )

{

    // Create a stack of capacity equal to size

    Stack* stack = createStack( size );

    // The first element of pre[] is always root

    Node* root = newNode( pre[0] );

    // Push root

    push( stack, root );

    int i;

    Node* temp;

    // Iterate through rest of the size-1 items of given preorder array

    for ( i = 1; i < size; ++i )

    {

        temp = NULL;

        /* Keep on popping while the next value is greater than

           stack's top value. */

        while ( !isEmpty( stack ) && pre[i] > peek( stack )->data )

            temp = pop( stack );

        // Make this greater value as the right child and push it to the stack

        if ( temp != NULL)

        {

            temp->right = newNode( pre[i] );

            push( stack, temp->right );

        }

        // If the next value is less than the stack's top value, make this value

        // as the left child of the stack's top node. Push the new node to stack

        else

        {

            peek( stack )->left = newNode( pre[i] );

            push( stack, peek( stack )->left );

        }

    }

    return root;

}

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