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Problem solution · Java

Binary Search Tree Iterator II

Binary Search Tree Iterator II: a Java solution using stack-based processing. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Stack-based processing
Source
walkccc LeetCode Solutions
Length
39 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Stack-based processing

For Binary Search Tree Iterator II, the implementation keeps unresolved items in last-in, first-out order, often to match boundaries, parse structure, or maintain monotonic candidates.

  1. Define what every stack entry represents.
  2. Pop entries once the current item resolves or invalidates them.
  3. Push the current item with only the information later steps need.

Code notes

  • 39 lines of Java from the credited upstream file 1586.java.
  • The implementation visibly relies on work queue.
  • 1 loop block detected.

Complexity

If each item is pushed and popped at most once, the stack work is linear.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeBinary Search Tree Iterator II · JavaJava
Use this to learn the idea, then write your own version.
class BSTIterator {  public BSTIterator(TreeNode root) {    pushLeftsUntilNull(root);  }   public boolean hasNext() {    return !nexts.isEmpty();  }   public int next() {    Pair<TreeNode, Boolean> pair = nexts.pop();    TreeNode root = pair.getKey();    boolean fromNext = pair.getValue();    if (fromNext)      pushLeftsUntilNull(root.right);    prevsAndCurr.push(root);    return root.val;  }   public boolean hasPrev() {    return prevsAndCurr.size() > 1;  }   public int prev() {    nexts.push(new Pair<>(prevsAndCurr.pop(), /*fromNext=*/false));    return prevsAndCurr.peek().val;  }   private void pushLeftsUntilNull(TreeNode root) {    while (root != null) {      nexts.push(new Pair<>(root, /*fromNext=*/true));      root = root.left;    }  }   private Deque<TreeNode> prevsAndCurr = new ArrayDeque<>();  private Deque<Pair<TreeNode, /*fromNext=*/Boolean>> nexts = new ArrayDeque<>();}

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