[leedcode 173] Binary Search Tree Iterator

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.

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

public class BSTIterator {
    Stack<TreeNode> stack=new Stack<TreeNode>();
    //中序遍历+stack
    //非递归中序遍历二叉树的话就是借助栈，让下次输出的目标节点始终存放在栈顶位置。每次输出一个节点，
    //就将此节点的后续节点放入栈中（沿着右子树的左子树一直向左就是下一个要输出的节点）。
    public BSTIterator(TreeNode root) {
        pushStack(root);
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.empty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode node=stack.pop();
        pushStack(node.right);///注意保存右节点
        return node.val;
    }
    private void pushStack(TreeNode node){
            while(node!=null){
                stack.push(node);
                node=node.left;
            }
    } 
}

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