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.
实现一个二叉搜索树迭代器 返回下一个最小值 容易想到binary search tree的inorder遍历 是有序的
想到了inorder 也想到了stack 就是递归没想好 下面是自己根据答案改写的
public class BSTIterator { private Stack<TreeNode> stack = new Stack<TreeNode>(); public BSTIterator(TreeNode root) { pushLeft(root); } private void pushLeft(TreeNode root) { if (root == null) return; stack.push(root); pushLeft(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 node = stack.pop(); pushLeft(node.right); return node.val; } }