173. Binary Search Tree Iterato

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.

思路：用stack来实现，将所有左子树push进stack。把next pop出来之后需要维护stack，检查右子树，如果存在继续push

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
Stack<TreeNode> res;
    public BSTIterator(TreeNode root) {
        res=new Stack<>();
        while(root!=null){
            res.push(root);
            root=root.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if(!res.isEmpty()){
            return true;
        }
        return false;
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode node=res.pop();
        int val=node.val;
        if(node.right!=null){
            node=node.right;
            while(node!=null){
                res.push(node);
                node=node.left;
            }
        }
        return val;
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */