# 173. 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.

# Solution

Approach 1: Stack.

# Code (Python)

Approach 1:

# Code (C++)

Approach 1:

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