Binary Search Tree Iterator

Binary Search Tree Iterator

Total Accepted: 34935 Total Submissions: 109369 Difficulty: 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.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

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

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    BSTIterator(TreeNode *root) {
        _cur = root;
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        while(_cur){
            paths.push(_cur);
            _cur = _cur->left;
        }
        return paths.empty() ? false : true;
    }

    /** @return the next smallest number */
    int next() {
        TreeNode* p = paths.top();
        paths.pop();
        _cur = p->right;
        return p->val;
    }
private:
    TreeNode* _cur;
    stack<TreeNode*> paths;
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

Next challenges: (M) Flatten 2D Vector (M) Zigzag Iterator (M) Peeking Iterator (M) Inorder Successor in BST