Binary Search Tree Iterator

Problem Statement

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.

Approach

• Push all the left nodes to the stack starting from the root in the constructor. This will push the nodes in descending order.
• In the hasNext() return true if the stack has elements and false if its empty
• In the next(), pop the node from the stack, push to the stack the right subtree of this node and return its value.

Run Time Complexity

hasNext(): O(1)

next(): O(logn)

memory: O(logn)

Code

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode next;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class BSTIterator {
    Stack<TreeNode> stack = new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
        pushToLeft(root);
    }
/** @return whether we have a next smallest number */
    public boolean hasNext(){
        if(stack.empty()){
            return false;
        }return true;
    }
/** @return the next smallest number */
    public int next(){
        TreeNode node = stack.pop();
        pushToLeft(node.right);
        return node.val;
    }
    
    private void pushToLeft(TreeNode node){
        if (node != null){
            stack.push(node);
            pushToLeft(node.left);
        }
    }
}
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */