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.
Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class BSTIterator { private Stack<TreeNode> stack=new Stack<TreeNode>(); public BSTIterator(TreeNode root) { pushToStack(root); } public void pushToStack(TreeNode root){ while(root != null){ stack.push(root); root = root.left; } } /** @return whether we have a next smallest number */ public boolean hasNext() { return !stack.isEmpty(); } /** @return the next smallest number */ public int next() { TreeNode cur = stack.pop(); pushToStack(cur.right); return cur.val; } } /** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */
public class BSTIterator { static ArrayList<Integer> a; int i; public BSTIterator(TreeNode root) { i = 0; a = new ArrayList<Integer>(); inorder(root, a); } public void inorder(TreeNode root, ArrayList<Integer> a) { if (root == null) { return; } inorder(root.left, a); a.add(root.val); inorder(root.right, a); } /** @return whether we have a next smallest number */ public boolean hasNext() { return (i < a.size()) ? true : false; } /** @return the next smallest number */ public int next() { return a.get(i++); } }