Leetcode 之 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.

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想了半天没想出来这道题的考点，是考全局变量吗？还是考BST中序遍历是递增序列？

中序遍历了一遍直接判断竟然就A了，连测试集都木有构建，真是史上前所未有。。

再回顾一下BST吧。小的值在左子树，大的数在右子树。这样的结构很容易拿到最值，即最左边的节点是最小的，最右节点是最大值。中序遍历可以得到一个ordered list。

上代码吧~

public class BSTIterator {    ArrayList<Integer> orderedList = new ArrayList<Integer>();    int index,length;    public void MiddleOrder(TreeNode root){        if(root != null){            MiddleOrder(root.left);            orderedList.add(root.val);            MiddleOrder(root.right);        }    }    public BSTIterator(TreeNode root) {        MiddleOrder(root);        index = 0;        length = orderedList.size();    }    /** @return whether we have a next smallest number */    public boolean hasNext() {        if(index < length){            index++;            return true;        }        return false;    }    /** @return the next smallest number */    public int next() {        return orderedList.get(index - 1);    }}

61 / 61 test cases passed.
Status: Accepted
Runtime: 404 ms