Binary Search Tree Iterator

最新推荐文章于 2024-04-09 22:21:57 发布

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本文介绍了一种二叉搜索树（BST）迭代器的实现方法，该迭代器可以在平均O(1)的时间复杂度内返回BST中的下一个最小元素，并使用O(h)的空间复杂度，其中h为树的高度。

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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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

tips:实现BST算法，根据以往的知识，BST使用队列实现的，Java中就是Queue和LinkedList实例

代码

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

public class BSTIterator {
   

    Queue<Integer> q=new LinkedList<Integer>();
    public BSTIterator(TreeNode root) {
       store(root);
    }


     public void store(TreeNode root){  
          
        if(root!=null){  
            store(root.left);  
            q.offer(root.val);  
            store(root.right);  
        }  
    }  
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return q.peek() != null;  
    }

    /** @return the next smallest number */
    public int next() {
        return q.poll();
    }
}

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