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BSTIterator.java
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BSTIterator.java
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package Leetcode;
import java.util.ArrayDeque;
import java.util.Deque;
/**
* @author kalpak
*
* 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?
*/
public class BSTIterator {
private TreeNode root;
private Deque<TreeNode> stack;
public BSTIterator(TreeNode root) {
this.root = root;
stack = new ArrayDeque<>();
buildInorderBST(root);
}
private void buildInorderBST(TreeNode root) {
while(root != null) {
stack.push(root);
root = root.left;
}
}
public int next() {
TreeNode current = stack.pop();
buildInorderBST(current.right);
return current.val;
}
public boolean hasNext() {
return !stack.isEmpty();
}
public static void main(String[] args) {
TreeNode root = new TreeNode(7);
root.left = new TreeNode(3);
root.right = new TreeNode(15);
root.right.left = new TreeNode(9);
root.right.right = new TreeNode(20);
BSTIterator obj = new BSTIterator(root);
System.out.println(obj.next());
System.out.println(obj.next());
System.out.println(obj.hasNext());
System.out.println(obj.next());
System.out.println(obj.hasNext());
System.out.println(obj.next());
System.out.println(obj.hasNext());
System.out.println(obj.next());
System.out.println(obj.hasNext());
}
}