package LeetCode_173
import java.util.*
/**
* 173. Binary Search Tree Iterator
* https://leetcode.com/problems/binary-search-tree-iterator/description/
*
* Implement an iterator over a binary search tree (BST).
* Your iterator will be initialized with the root node of a BST.
Calling next() will return the next smallest number in the BST.
Note:
next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
You may assume that next() call will always be valid, that is,
there will be at least a next smallest number in the BST when next() is called.
* */
class TreeNode(var `val`: Int) {
var left: TreeNode? = null
var right: TreeNode? = null
}
/*
* solution 1: in-order + ArrayList, But need O(n) space
* solution 2: Stack, O(1) time and O(h) space in next() and hasNext()
* */
class BSTIterator(root: TreeNode?) {
val stack = Stack<TreeNode>()
init {
setNext(root)
}
/** @return the next smallest number */
fun next(): Int {
val node = stack.pop()
//set next for right node
setNext(node.right)
return node.`val`
}
/** @return whether we have a next smallest number */
fun hasNext(): Boolean {
return stack.isNotEmpty()
}
private fun setNext(root_: TreeNode?) {
//push the left node in stack, because left node in BST is always small
//stack: first in last out
var root = root_
while (root != null) {
stack.push(root)
root = root.left
}
}
}
/**
* Your BSTIterator object will be instantiated and called as such:
* var obj = BSTIterator(root)
* var param_1 = obj.next()
* var param_2 = obj.hasNext()
*/