Binary Search Tree Iterator

Design an iterator over a binary search tree with the following rules:

• Elements are visited in ascending order (i.e. an in-order traversal)
• next() and hasNext() queries run in O(1) time in average.

Example

For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]

   10
 /    \
1      11
 \       \
  6       12

Challenge 

Extra memory usage O(h), h is the height of the tree.

Super Star: Extra memory usage O(1)

 

Analyse: Recursion.

Runtime: 25ms

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 * Example of iterate a tree:
 * BSTIterator iterator = BSTIterator(root);
 * while (iterator.hasNext()) {
 *    TreeNode * node = iterator.next();
 *    do something for node
 */
class BSTIterator {
public:
    //@param root: The root of binary tree.
    BSTIterator(TreeNode *root) {
        // write your code here
        inorder = inorderTraversal(root);
        index = 0;
    }

    //@return: True if there has next node, or false
    bool hasNext() {
        // write your code here
        return index < inorder.size();
    }
    
    //@return: return next node
    TreeNode* next() {
        // write your code here
        if (index >= inorder.size()) return NULL;
        return inorder[index++];
    }
    
private:
    vector<TreeNode* > inorder;
    int index;
    
    vector<TreeNode* > inorderTraversal(TreeNode* root) {
        vector<TreeNode* > result;
        
        helper(result, root);
        return result;
    }
    
    void helper(vector<TreeNode* >& result, TreeNode* root) {
        if (!root) return;
        
        helper(result, root->left);
        result.push_back(root);
        helper(result, root->right);
    }
};