The O(n) Club: Binary Search Tree Iterator: Lazy Traversal for Stack-Savvy Devs

The O(n) Club: Binary Search Tree Iterator: Lazy Traversal for Stack-Savvy Devs

⚡ TL;DR

You need to traverse a BST in-order, get the next element on demand, and—here’s the kicker—do it without stuffing your RAM like it’s a Black Friday shopping cart. Stay smooth, use a stack…or just copy this Java snippet and bask in the glory:

// Java, but make it quick and stacky
TreeNode root = ...;
BSTIterator iter = new BSTIterator(root);
while (iter.hasNext()) {
    System.out.println(iter.next());
}

🧠 How Devs Usually Mess This Up

There are two kinds of engineers: those who precompute everything (O(n) arrays, because apparently you love wasting RAM), and those who try to build a web of pointers worthy of a family drama. Spoiler: neither gets you a raise—or the job.

• Eager List People: You jam the whole in-order sequence into a list first. Congrats on turning your beautiful BST into a sad, space-hogging array.
• Pointer Acrobats: You think you need next, prev, uncle, and grandmother pointers. Sorry, LeetCode only needs forward now. Let’s not overengineer our pain, mmkay?

🔍 What’s Actually Going On

Imagine you’re a robot chef hunting down the smallest ingredient in a pantry. Every time you want the “next” smallest, you walk as far left as possible (because that’s where the fresh basil is…obviously). You push each pantry door (node) you pass onto a stack, so you can retrace your steps. When you grab an ingredient, check if there’s a right shelf (right child); if so, start stacking left again. The stack remembers your journey—so you never repeat steps, never peek ahead, and definitely don’t take everything out of the pantry at once. This is lazy, casual, and O(h) space efficient—just like every Friday dev meeting should be.

🛠️ PseudoCode

• Initialize: Stack up all left descendants from root.

  void pushLeft(TreeNode node) {
      while (node != null) {
          stack.push(node);
          node = node.left;
      }
  }

  The smallest element is now at the top. You overachiever.
• hasNext(): Returns true if you have unfinished business.

  boolean hasNext() {
      return !stack.isEmpty();
  }

  The coffee test: If the stack isn’t empty, keep going.
• next(): Pop the smallest, then push the left path of its right child (if any).

  int next() {
      TreeNode curr = stack.pop();
      if (curr.right != null) {
          pushLeft(curr.right);
      }
      return curr.val;
  }

  Like unboxing Russian dolls. Only with trees.

💻 The Code

class BSTIterator {
    private Stack
<treenode> stack = new Stack<>();
     public BSTIterator(TreeNode root) {
        pushLeft(root);
    }
     public boolean hasNext() {
        return !stack.isEmpty();
    }
     public int next() {
        TreeNode node = stack.pop();
        if (node.right != null) {
            pushLeft(node.right);
        }
        return node.val;
    }
     private void pushLeft(TreeNode node) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }
}
 class TreeNode {
    int val;
    TreeNode left, right;
    TreeNode(int x) { val = x; }
}</treenode>

⚠️ Pitfalls, Traps & Runtime Smacks

• Packing the Whole Tree: Don’t precompute everything, unless you’re looking for a performance disaster. Stick to O(h) space—that’s tree height, not the number of nodes.
• A Stack Is Not Infinite: If your BST is actually a glorified LinkedList (aka completely unbalanced), worst-case stack size is as bad as your last code review.
• No Multithreading Shenanigans: Your iterator is as thread-safe as juggling chainsaws—so steer clear of concurrency unless you like chaos.
• Amortized O(1): Yes, next() and hasNext() aren’t always constant, but on average they totally are. That counts for interviews, or so the rumor goes.

🧠 Interviewer Brain-Tattoo

Only stack what you need, when you need it. Don’t binge-traverse. The BST iterator isn’t just lazy—it’s professionally lazy. (And if your solution uses O(n) memory anyway, do everyone a favor and call it a TreeArrayator.)