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Iterator for a Binary Search Tree

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 12

Write a C# program to implement an iterator for a binary search tree (BST) with the following operations: 1. BSTIterator(TreeNode root): Initialize an iterator with the root of a BST. 2. bool HasNext(): Return true if there is a next element in the iteration. 3. int Next(): Return the next smallest element in the BST. The iterator should follow the in-order traversal of the BST (left-root-right).

Example 1
    
  
  • Input: root = [7, 3, 15, null, null, 9, 20]
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  • Operations: 
    
      
      
    • bSTIterator = new BSTIterator(root);
      • 
      
    • bSTIterator.Next();    // return 3
      • 
      
    • bSTIterator.Next();    // return 7
      • 
      
    • bSTIterator.HasNext(); // return true
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    • bSTIterator.Next();    // return 9
      • 
      
    • bSTIterator.Next();    // return 15
      • 
      
    • bSTIterator.Next();    // return 20
      • 
      
    • bSTIterator.HasNext(); // return false
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Example 2
    
  
  • Input: root = [1, null, 3, null, null, 2, null]
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  • Operations: 
    
      
      
    • bSTIterator = new BSTIterator(root);
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    • bSTIterator.HasNext(); // return true
      • 
      
    • bSTIterator.Next();    // return 1
      • 
      
    • bSTIterator.Next();    // return 2
      • 
      
    • bSTIterator.Next();    // return 3
      • 
      
    • bSTIterator.HasNext(); // return false
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Constraints
    
  
  • The number of nodes in the tree is in the range [1, 10^5]
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  • 0 ≤ Node.val ≤ 10^6
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  • At most 10^5 calls will be made to HasNext and Next
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  • Time Complexity: O(1) average time per operation
    • 
  
  • Space Complexity: O(h) where h is the height of the tree
    •
TreeStackGoldman SachsD. E. Shaw
Editorial

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Hint
Solution

Solution Hints

    
  
  • Use a stack to keep track of nodes
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  • Initialize the stack by pushing all leftmost nodes
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  • To get the next smallest element, pop the top of the stack
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  • After popping a node, push all leftmost nodes of its right child
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  • The HasNext method checks if the stack is empty
    •

Steps to solve by this approach:

 Step 1: Create a stack to store the nodes for in-order traversal.

 Step 2: Implement a helper method (PushAllLeft) to push all leftmost nodes of a subtree onto the stack.

 Step 3: In the constructor, initialize the stack and push all leftmost nodes of the root.

 Step 4: For the Next() method, pop the top node from the stack.

 Step 5: If the popped node has a right child, push all leftmost nodes of the right child onto the stack.

 Step 6: Return the value of the popped node.

 Step 7: For the HasNext() method, simply check if the stack has any nodes left.

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