leetcode--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, whereh is the height of the tree.

题意：给定一棵BST树。创建一个next()函数可以返回下一个最小的节点值，创建一个hasNext()函数判断是否还有下一个节点。

要求next()的时间复杂度为O(n),空间复杂度为O(h)

解法1：中序遍历二叉树，保留遍历结果。

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

public class BSTIterator {
ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
int cur = -1;
int size = 0;
public BSTIterator(TreeNode root) {
if(root!=null) helper(root);
size = queue.size();
}

void helper(TreeNode root){
if(root.left!=null) helper(root.left);
queue.add(root);
if(root.right!=null) helper(root.right);
}

/** @return whether we have a next smallest number */
public boolean hasNext() {
return cur<size-1;
}

/** @return the next smallest number */
public int next() {
cur++;
return queue.get(cur).val;
}
}

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