LeetCode 173. Binary Search Tree Iterator

Question

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

• BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
• boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
• int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

https://leetcode.com/problems/binary-search-tree-iterator/

Solution

• Solution1 – inorder

class BSTIterator {
    Stack<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        pushAllLeft(root);
    }
    
    public int next() {
        TreeNode node = stack.pop();
        pushAllLeft(node.right);
        return node.val;
    }
    
    public boolean hasNext() {
        return !stack.isEmpty();
    }
    public void pushAllLeft(TreeNode root){
        while(root != null){
            stack.push(root);
            root = root.left;
        }
    }
}

Complexity:

Time complexity: O(h)

Space complexity: O(h)