Binary Search Tree Iterator

Description

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.

Have you met this question in a real interview?   Yes

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)

solution1: update as binary list

/**
 * 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) {
        // do intialization if necessary
        TreeNode * parentroot = NULL;
        TreeToBinaryList(root, parentroot);
        while(root != NULL && root->left != NULL) {
            root = root->left;
        }
        start = root;
    }

    /*
     * @return: True if there has next node, or false
     */
    bool hasNext() {
        // write your code here
        return start != NULL;
    }

    /*
     * @return: return next node
     */
    TreeNode * next() {
        // write your code here
        TreeNode * tmp = start;
        start = start->right;
        return tmp;
    }
private:
    void TreeToBinaryList(TreeNode * root, TreeNode *& parentroot) {
        if (root == NULL) {
            return;
        }
        TreeToBinaryList(root->left, parentroot);
        root->left = parentroot;
        if (parentroot != NULL) {
            parentroot->right = root;
        }
        parentroot = root;
        TreeToBinaryList(root->right, parentroot);
    }
    TreeNode * start;
};

solution2: use stack

/**
 * 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) {
        // do intialization if necessary
        while (root != NULL) {
            st.push(root);
            root = root->left;
        }
    }

    /*
     * @return: True if there has next node, or false
     */
    bool hasNext() {
        // write your code here
        return st.empty() == false;
    }

    /*
     * @return: return next node
     */
    TreeNode * next() {
        // write your code here
        TreeNode * tmp = st.top();
        st.pop();
        TreeNode * root = tmp->right;
        while (root != NULL) {
            st.push(root);
            root = root->left;
        }
        return tmp;
    }
private:
    stack st;
};