LeetCode-173-Binary Search Tree Iterator

最新推荐文章于 2022-06-10 17:57:38 发布

转载最新推荐文章于 2022-06-10 17:57:38 发布 · 115 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/nobodywang/p/10400307.html

本文介绍了一种在二叉搜索树中实现迭代器的方法，该迭代器可以在平均O(1)的时间复杂度和O(h)的空间复杂度下运行next()和hasNext()操作，其中h为树的高度。通过使用栈来模拟先序遍历，确保每次调用next()都能返回树中下一个最小的数值。

算法描述：

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.

Example:

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

Note:

next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

解题思路：用栈模拟先序遍历。

public:
    BSTIterator(TreeNode* root) {
        while(root!=nullptr){
            stk.push(root);
            root=root->left;
        }
    }
    
    /** @return the next smallest number */
    int next() {
        int res = -1;
        if(!stk.empty()){
            TreeNode* cur = stk.top();
            stk.pop();
            res = cur->val;
            cur=cur->right;
            while(cur!=nullptr){
                stk.push(cur);
                cur=cur->left;
            }
        }
        return res;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !stk.empty();
    }
private:
    stack<TreeNode*> stk;

转载于:https://www.cnblogs.com/nobodywang/p/10400307.html