173. Binary Search Tree Iterator (Medium)


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.



BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false



  • next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
  • You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.



class BSTIterator {
    Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<>();

        // left section is always smaller, visit left at first

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

    public int next() {
        TreeNode node = stack.pop();
        int result = node.val;

        // 'node' will be visited only if its left children nodes are all visited
        // since 'node' visited, then visit its right children nodes
        if (node.right != null) {
            node = node.right;


        return result;

    public void pushLeft(TreeNode node) {
        while (node != null) {
            node = node.left;

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