Binary Search Tree Iterator — LeetCode 173 Python Solution

The problem

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.

Example

Input

["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]

Output

[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation

BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);

Python solution

# 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):
        def inorder(root):
            if root:
                inorder(root.left)
                self.vals.append(root.val)
                inorder(root.right)

        self.cur = 0
        self.vals = []
        inorder(root)

    def next(self) -> int:
        res = self.vals[self.cur]
        self.cur += 1
        return res

    def hasNext(self) -> bool:
        return self.cur < len(self.vals)


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

Complexity

Pattern: Stack

When the most recent unresolved thing is the one that matters, use a stack. LeetCode 173. Binary Search Tree Iterator is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Stack.

The stack guide has the Python template for the pattern and the 194 LeetCode problems that use it.

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