Binary Search Tree Iterator

问题描述

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.

 

解决思路

中序遍历+非递归（辅助栈）

时间/空间复杂度均为O(logn).

 

程序

public class BSTIterator {

    private Stack<TreeNode> s;

    

    public BSTIterator(TreeNode root) {

        s = new Stack<>();

        pushLeft(root);

    }

    

    private void pushLeft(TreeNode node) {

        TreeNode cur = node;

        while (cur != null) {

            s.push(cur);

            cur = cur.left;

        }

    }



    /** @return whether we have a next smallest number */

    public boolean hasNext() {

        return !s.isEmpty();

    }



    /** @return the next smallest number */

    public int next() {

        TreeNode node = s.pop();

        pushLeft(node.right);

        return node.val;

    }

}