173.binary-search-tree-iterator.cpp

//  Tag: Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator
//  Time: O(1)
//  Space: O(H)
//  Ref: -
//  Note: InOrder

//  Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):
//  
//  BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
//  boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
//  int next() Moves the pointer to the right, then returns the number at the pointer.
//  
//  Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.
//  You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.
//   
//  Example 1:
//  
//  
//  Input
//  ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
//  [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
//  Output
//  [null, 3, 7, true, 9, true, 15, true, 20, false]
//  
//  Explanation
//  BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
//  bSTIterator.next();    // return 3
//  bSTIterator.next();    // return 7
//  bSTIterator.hasNext(); // return True
//  bSTIterator.next();    // return 9
//  bSTIterator.hasNext(); // return True
//  bSTIterator.next();    // return 15
//  bSTIterator.hasNext(); // return True
//  bSTIterator.next();    // return 20
//  bSTIterator.hasNext(); // return False
//  
//   
//  Constraints:
//  
//  The number of nodes in the tree is in the range [1, 105].
//  0 <= Node.val <= 106
//  At most 105 calls will be made to hasNext, and next.
//  
//   
//  Follow up:
//  
//  Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?
//  
//  

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
public:
    TreeNode *cur;
    stack<TreeNode *> st;
    BSTIterator(TreeNode* root) {
        cur = root;
    }
    
    int next() {
        while (cur) {
            st.push(cur);
            cur = cur->left;
        }

        cur = st.top();
        st.pop();
        int res = cur->val;
        cur = cur->right;
        return res;
    }
    
    bool hasNext() {
        return (cur || !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();
 */