Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
Callingnext()will return the next smallest number in the BST.
Note:next()andhasNext()should run in average O(1) time and uses O(h) memory, wherehis the height of the tree.
Solution
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
private final Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
stack = new Stack<>();
TreeNode cur = root;
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode node = stack.pop();
// Traversal cur node's right branch
TreeNode cur = node.right;
while (cur != null){
stack.push(cur);
cur = cur.left;
}
return node.val;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/