[leetcode]Binary Search Tree Iterator

最新推荐文章于 2024-04-09 22:21:57 发布

雨中鹰

最新推荐文章于 2024-04-09 22:21:57 发布

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分类专栏： leetcode 文章标签： leetcode Binary Search Tree I

本文链接：https://blog.youkuaiyun.com/menglinaoxiang/article/details/45101863

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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，该迭代器可在平均O(1)时间内返回BST中的下一个最小值，并使用O(h)内存，其中h为树的高度。文章详细阐述了如何通过辅助栈来跟踪节点，确保高效地遍历整个BST。

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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.

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

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
private:
    stack<TreeNode*> st; //辅助栈
    TreeNode *curNode; //当前的元素
public:
    BSTIterator(TreeNode *root) {
        curNode = NULL;
        //左枝入栈
        while(root != NULL){
            st.push(root);
            root = root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        if(curNode == NULL){ //取出最小的
            if(!st.empty()){
                curNode = st.top(); st.pop();
                return true;
            }
            return false;
        }
        
        TreeNode *node = curNode->right;
        if(node == NULL){ //右子树为空， 则继续从栈中取出
            if(!st.empty()){
                curNode = st.top(); st.pop();
                return true;
            }
            return false;           
        }
        if(node->left == NULL){ 
            //右子树不空， 检查该右子树的左子树是否空，如果空直接返回当前右子树
            curNode = node;
            return true;
        }
        while(node->left){ //右子树不空， 当前右子树的左子树不空，入栈
            st.push(node);
            node = node->left;
            curNode = node;
        }
        return true;
    }

    /** @return the next smallest number */
    int next() {
        return curNode->val;
    }
};

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