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 com_github_leetcode.TreeNode;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class BSTIterator {
private TreeNode node;
public BSTIterator(TreeNode root) {
node = root;
}
public int next() {
int res = -1;
while (node != null) {
if (node.left != null) {
TreeNode rightMost = node.left;
while (rightMost.right != null) {
rightMost = rightMost.right;
}
rightMost.right = node;
TreeNode temp = node.left;
node.left = null;
node = temp;
} else {
res = node.val;
node = node.right;
return res;
}
}
return res;
}
public boolean hasNext() {
return node != null;
}
}
/*
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator obj = new BSTIterator(root);
* int param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/