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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.
Tree Stack
This is an extremely important question, if you are going for an interview. I repeat: this is an extremely important question, if you are going for an interview. If you do not remember it by heart, I will repeat again.
The solution of the iterator applies to a lot of related questions. So make sure you practise this question until you are perfect. You WILL BE ASKED this question at times.
You could read my other post [Question] Iterator of Binary Search Tree.
We only need to keep 1 variable in RAM, that is a stack.
public class BSTIterator { Stack<TreeNode> stack = new Stack<TreeNode>(); public BSTIterator(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() { if (!hasNext()) { return 0; } TreeNode next = stack.pop(); TreeNode node = next.right; while (node != null) { stack.push(node); node = node.left; } return next.val; } }
8.10 [LeetCode 173] Binary Search Tree Iterator
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原文地址：http://www.cnblogs.com/michael-du/p/4719786.html
