leetcode之Binary Search Tree Iterator

最新推荐文章于 2025-08-13 21:50:19 发布

luckyu1

最新推荐文章于 2025-08-13 21:50:19 发布

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文章标签： leetcode 算法面试

本文链接：https://blog.youkuaiyun.com/luckyu1/article/details/50754136

本文介绍如何使用迭代方式实现二叉搜索树（BST）的中序遍历，通过栈来满足平均O(1)时间复杂度和O(h)内存使用的约束。

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题目：

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.

解答：

由于BST的中序遍历就是从小到大的顺序，所以直接进行中序遍历即可，为了满足复杂度要求，需要将递归改为迭代过程，用栈实现

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    BSTIterator(TreeNode *root) {
        if(!root)
            return;
        while(root)
        {
            st.push(root);
            root = root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !st.empty();
    }

    /** @return the next smallest number */
    int next() {
        if(BSTIterator :: hasNext())
        {
        int res = st.top()->val;
        TreeNode *tmp = st.top();
        st.pop();
        if(tmp->right)
        {   
            tmp = tmp->right;
            while(tmp)
            {
                st.push(tmp);
                tmp = tmp->left;
            }
        }
        return res;
        }
    }
private:
    stack<TreeNode*> st;
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */