ID | Title | Difficulty | |
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173. Binary Search Tree Iterator
Medium
LeetCode
Stack, Tree, Design, Binary Search Tree, Binary Tree, Iterator
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 theBSTIterator
class. Theroot
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()
Returnstrue
if there exists a number in the traversal to the right of the pointer, otherwise returnsfalse
.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.
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 tohasNext
, andnext
.
Code
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
private TreeNode cur;
private Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
cur = root;
stack = new Stack<>();
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
if(!stack.isEmpty() || cur != null) {
return true;
} else {
return false;
}
}
/** @return the next smallest number */
public int next() {
while(cur != null){
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
int val = cur.val;
cur = cur.right;
return val;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
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