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.

Test Cases

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?

Solution

        
/**
 * 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 {
    queue<int> q;
    void inOrder(TreeNode* root, queue<int> &q) {
        if(root == NULL){
            return;
        }
        inOrder(root->left, q);
        q.push(root->val);
        inOrder(root->right, q);
    }
public:
    BSTIterator(TreeNode* root) {
        inOrder(root, q);
    }

    int next() {
        if(!q.empty()){
            int frontNode = q.front();
            q.pop();
            return frontNode;
        }
        return 0;
    }

    bool hasNext() {
        if(!q.empty()){
            return true;
        }
        return false;
    }
};

/**
 * 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();
 */

    

        
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class BSTIterator {
    private Deque<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new ArrayDeque<>();
        TreeNode next = root;
        while(next != null) {
            stack.offerLast(next);
            next = next.left;
        }
    }

    public int next() {
        TreeNode removedNode = stack.pollLast();
        TreeNode next = removedNode.right;
        while(next != null) {
            stack.offerLast(next);
            next = next.left;
        }
        return removedNode.val;
    }

    public boolean hasNext() {
        return !stack.isEmpty();

    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

    

        
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class BSTIterator:

    def __init__(self, root: TreeNode):
        self.lst = []
        self.index = 0
        if root:
            st = [root]
            while len(st) > 0:
                x = st.pop()
                self.lst.append(x.val)
                if x.left: st.append(x.left)
                if x.right: st.append(x.right)
            self.lst = sorted(self.lst)
            print(self.lst)

    def next(self) -> int:
        """
        @return the next smallest number
        """
        x = self.lst[self.index]
        self.index += 1
        return x

    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        if len(self.lst) == 0: return False
        if self.index < len(self.lst):
            return True
        return False

# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

    

Binary Search Tree Iterator

Topics

Companies

Platforms

Test Cases

Solution

Time Complexity: O(n)

Space Complexity: O(n)