[leetcode]Binary Search Tree Iterator

最新推荐文章于 2025-02-03 15:37:10 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，该迭代器能够以平均O(1)的时间复杂度返回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。通过利用栈结构存储节点，可以有效地按顺序遍历BST。

https://leetcode.com/problems/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 = null;

    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        while(root != null) {
            stack.push(root);
            root = root.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
		TreeNode p = stack.pop();
		int next = p.val;
		p = p.right;
		while (p != null) {
			stack.push(p);
			p = p.left;
		}
		return next;
    }
}

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