Binary Search Tree Iterator

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Solution:

  1. /**
  2.  * Definition for binary tree
  3.  * public class TreeNode {
  4.  *     int val;
  5.  *     TreeNode left;
  6.  *     TreeNode right;
  7.  *     TreeNode(int x) { val = x; }
  8.  * }
  9.  */
  11. public class BSTIterator {
  12.   Stack<TreeNode> stack = null;
  13.   TreeNode current = null;
  15.   public BSTIterator(TreeNode root) {
  16.     current = root;
  17.     stack = new Stack<TreeNode>();
  18.   }
  20.   /** @return whether we have a next smallest number */
  21.   public boolean hasNext() {
  22.     return !stack.isEmpty() || current != null;
  23.   }
  25.   /** @return the next smallest number */
  26.   public int next() {
  27.     while (current != null) {
  28.       stack.push(current);
  29.       current = current.left;
  30.     }
  32.     TreeNode t = stack.pop();
  33.     current = t.right;
  34.     return t.val;
  35.   }
  36. }
  38. /**
  39.  * Your BSTIterator will be called like this:
  40.  * BSTIterator i = new BSTIterator(root);
  41.  * while (i.hasNext()) v[f()] = i.next();
  42.  */