173. Binary Search Tree Iterator

173. Binary Search Tree Iterator (Medium)

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.

Companies:
Facebook, Amazon, Microsoft, Oracle, Bloomberg, Qualtrics, Cloudera, Uber, Google, LinkedIn, Cisco

Related Topics:
Stack, Tree, Design

Similar Questions:

• Binary Tree Inorder Traversal (Medium)
• Flatten 2D Vector (Medium)
• Zigzag Iterator (Medium)
• Peeking Iterator (Medium)
• Inorder Successor in BST (Medium)

Solution 1. Stack

// OJ: https://leetcode.com/problems/binary-search-tree-iterator/
// Author: github.com/lzl124631x
// Time: O(1) amortized
// Space: O(H)
class BSTIterator {
private:
    stack<TreeNode*> s;
    void pushNodes(TreeNode *node) {
        while (node) {
            s.push(node);
            node = node->left;
        }
    }
public:
    BSTIterator(TreeNode* root) {
        pushNodes(root);
    }
    int next() {
        auto node = s.top();
        s.pop();
        pushNodes(node->right);
        return node->val;
    }
    bool hasNext() {
        return s.size();
    }
};