173. Binary Search Tree Iterator

Description

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.

• 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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input:

["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext

[[[7, 3, 15, null, null, 9, 20]], [null], [null], [null], [null], [null], [null], [null], [null], [null]]

Output:

[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation:

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

/**
 * 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();
 */
 
class BSTIterator {
    private Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        pushNodes(root);
    }
    
    /** @return the next smallest number */
    public int next() {
        TreeNode root = stack.pop();
        pushNodes(root.right);
        return root.val;
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }
    
    private void pushNodes(TreeNode root) {
        while(root != null) {
            stack.push(root);
            root = root.left;
        }
    }
}

Follow up