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[leetcode] Binary Search Tree
题目：（Tree Stack）
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.
题解：
tree的题很容易就和stack联系在一起了，这道题就是维护一个stack就可以。
用pushleft()
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class BSTIterator { //TreeNode root; Stack<TreeNode> stack = new Stack<TreeNode>(); public BSTIterator(TreeNode root) { pushLeft(root); } /** @return whether we have a next smallest number */ public boolean hasNext() { return !stack.isEmpty(); } /** @return the next smallest number */ public int next() { TreeNode small=stack.pop(); pushLeft(small.right); return small.val; } public void pushLeft(TreeNode root) { while(root!=null) { stack.push(root); root=root.left; } }}/** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i.next(); */
不断地将root.left push 到stack，当回溯的时候如果当时那个node有right child，那么多那个right child 再进行pushleft()。
[leetcode] Binary Search Tree
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