Question
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.
For example: Given the binary search tree,
7
/ \
3 15
/ \
9 20
BSTIterator iterator = new BSTIterator(root);
iterator.next(); // return 3
iterator.next(); // return 7
iterator.hasNext(); // return true
iterator.next(); // return 9
iterator.hasNext(); // return true
iterator.next(); // return 15
iterator.hasNext(); // return true
iterator.next(); // return 20
iterator.hasNext(); // return false
Note:
next()
andhasNext()
should run inaverage O(1)
time and usesO(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 whennext()
is called.
Solution
TODO
Use a stack do inorder traversal.
Code
Here is a sample solution.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class BSTIterator {
Stack<TreeNode> s;
public BSTIterator(TreeNode root) {
s = new Stack<TreeNode>();
collectLeftChildren(root);
}
/** @return the next smallest number */
public int next() {
TreeNode n = s.pop();
collectLeftChildren(n.right);
return n.val;
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !s.isEmpty();
}
public void collectLeftChildren(TreeNode node) {
while(node != null) {
s.push(node);
node = node.left;
}
}
}
/**
* 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();
*/
Performance
next()
and hasNext()
should run in average O(1)
time and uses O(h)
memory, where h is the height of the tree.
-
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