LeetCode in TypeScript

173. Binary Search Tree Iterator

Medium

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?

Solution

import { TreeNode } from '../../com_github_leetcode/treenode'

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */
class BSTIterator {
    private node: TreeNode | null

    constructor(root: TreeNode | null) {
        this.node = root
    }

    next(): number {
        let res = -1
        while (this.node !== null) {
            if (this.node.left !== null) {
                let rightMost = this.node.left
                while (rightMost.right !== null && rightMost.right !== this.node) {
                    rightMost = rightMost.right
                }

                if (rightMost.right === null) {
                    rightMost.right = this.node
                    this.node = this.node.left
                } else {
                    rightMost.right = null
                    res = this.node.val
                    this.node = this.node.right
                    return res
                }
            } else {
                res = this.node.val
                this.node = this.node.right
                return res
            }
        }
        return res
    }

    hasNext(): boolean {
        return this.node !== null
    }
}

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

export { BSTIterator }

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