Last updated on October 5th, 2024 at 05:27 pm

Here, We see Binary Search Tree Iterator LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Binary Search Tree Iterator LeetCode Solution

Problem Statement

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

1. Binary Search Tree Iterator LeetCode Solution C++

class BSTIterator {
    stack<TreeNode *> myStack;
public:
    BSTIterator(TreeNode *root) {
        pushAll(root);
    }

    bool hasNext() {
        return !myStack.empty();
    }
    
    int next() {
        TreeNode *tmpNode = myStack.top();
        myStack.pop();
        pushAll(tmpNode->right);
        return tmpNode->val;
    }

private:
    void pushAll(TreeNode *node) {
        for (; node != NULL; myStack.push(node), node = node->left);
    }
};

2. Binary Search Tree Iterator LeetCode Solution Java

class BSTIterator {
    private Stack<TreeNode> stack = new Stack<TreeNode>();
    public BSTIterator(TreeNode root) {
        pushAll(root);
    }

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

    public int next() {
        TreeNode tmpNode = stack.pop();
        pushAll(tmpNode.right);
        return tmpNode.val;
    }
    
    private void pushAll(TreeNode node) {
        for (; node != null; stack.push(node), node = node.left);
    }
}

3. Binary Search Tree Iterator Solution JavaScript

var BSTIterator = function(root) {
    if(!root){
        this.data = [];
        return;
    }
    let visited = [];
    function traverse(node){
        if(node.left) traverse(node.left);
        visited.push(node.val);
        if(node.right) traverse(node.right);
    }
    traverse(root);
    this.data = visited.reverse();
};

BSTIterator.prototype.next = function() {
    return this.data.pop();
};

BSTIterator.prototype.hasNext = function() {
    return this.data.length > 0 ;
};

4. Binary Search Tree Iterator Solution Python

class BSTIterator(object):
    def __init__(self, root):
        self.stack = list()
        self.pushAll(root)

    def hasNext(self):
        return self.stack

    def next(self):
        tmpNode = self.stack.pop()
        self.pushAll(tmpNode.right)
        return tmpNode.val
        
    def pushAll(self, node):
        while node is not None:
            self.stack.append(node)
            node = node.left