173. Binary Search Tree Iterator (Medium)

https://leetcode.com/problems/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.

 

Example:

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

 

Note:

  • 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.

Hints

Solutions

https://www.programcreek.com/2014/04/leetcode-binary-search-tree-iterator-java/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class BSTIterator {
    Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }

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

    public int next() {
        TreeNode node = stack.pop();
        int result = node.val;
        if (node.right != null) {
            node = node.right;
            while (node != null) {
                stack.push(node);
                node = node.left;
            }
        }
        return result;
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */
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