173. Binary Search Tree Iterator

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

Problem

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.

Solution

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
    stack<TreeNode*> nodes;
    TreeNode* current;
public:
    BSTIterator(TreeNode* root) : current(nullptr) {
        while (root != nullptr) {
            nodes.push(root);
            root = root->left;
        }
    }
    
    /** @return the next smallest number */
    int next() {
        while (current != nullptr) {
            nodes.push(current);
            current = current->left;
        }
        TreeNode* next = nodes.top();
        nodes.pop();
        current = next->right;
        return next->val;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return (current != nullptr || !nodes.empty());
    }
};

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

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class BSTIterator:

    def __init__(self, root: TreeNode):
        self.current = None
        self.nodes = list()
        while root:
            self.nodes.append(root)
            root = root.left
            
    def next(self) -> int:
        """
        @return the next smallest number
        """
        while self.current:
            self.nodes.append(self.current)
            self.current = self.current.left
        next_el = self.nodes.pop()
        self.current = next_el.right
        return next_el.val
        

    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        return self.current or self.nodes
        


# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

• #tree

• #bst

• #inorder