365天挑战LeetCode1000题——Day 074 验证栈序列二叉搜索树中第K小的元素二叉搜索树迭代器-优快云博客

本文链接：https://blog.youkuaiyun.com/ShowMeTheCod3/article/details/126619056

本文介绍三种算法：验证栈序列是否有效的方法、在二叉搜索树中寻找第K小元素的策略及实现二叉搜索树迭代器的设计。通过具体代码实现，帮助读者深入理解这些算法的工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

946. 验证栈序列

在这里插入图片描述

代码实现（自解）

class Solution {
public:
    bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
        int n = pushed.size();
        int pp = 0;
        stack<int> myStack;
        for (int i = 0; i < n; i++) {
            myStack.push(pushed[i]);
            while (!myStack.empty() && myStack.top() == popped[pp]) {
                myStack.pop();
                pp++;
            }
       }
       return pp == n && myStack.empty();
    }
};

230. 二叉搜索树中第K小的元素

在这里插入图片描述

代码实现（自解）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        vector<int> arr;
        queue<TreeNode*> myQueue;
        myQueue.push(root);
        while (!myQueue.empty()) {
            int sz = myQueue.size();
            while (sz--) {
                TreeNode* tmp = myQueue.front();
                myQueue.pop();
                arr.push_back(tmp->val);
                if (tmp->left) myQueue.push(tmp->left);
                if (tmp->right) myQueue.push(tmp->right);
            }
        }
        sort(arr.begin(), arr.end());
        return arr[k - 1];
    }
};

173. 二叉搜索树迭代器

在这里插入图片描述

代码实现（自解）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
private:
    vector<int> iterator;
    int pi;
public:
    BSTIterator(TreeNode* root) {
        stack<TreeNode*> myStack;
        pi = 0;
        while (root || !myStack.empty()) {
            while (root) {
                myStack.push(root);
                root = root->left;
            }
            root = myStack.top();
            iterator.push_back(root->val);
            myStack.pop();
            root = root->right;
        }
    }
    
    int next() {
        return iterator[pi++];
    }
    
    bool hasNext() {
        return pi != iterator.size();
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */