代码随想录算法训练营第二十天 | 530.二叉搜索树的最小绝对差 501.二叉搜索树中的众数 236. 二叉树的最近公共祖先

530.二叉搜索树的最小绝对差

题目链接：https://leetcode.cn/problems/minimum-absolute-difference-in-bst/

文章讲解：https://programmercarl.com/0530.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E7%9A%84%E6%9C%80%E5%B0%8F%E7%BB%9D%E5%AF%B9%E5%B7%AE.html

视频讲解：https://www.bilibili.com/video/BV1DD4y11779/?vd_source=5287cb6df25410935566fc8f0d951a5a

//方法一
class Solution {
private:
vector<int> vec;
void traversal(TreeNode* root) {
    if (root == NULL) return;
    traversal(root->left);
    vec.push_back(root->val); // 将二叉搜索树转换为有序数组
    traversal(root->right);
}
public:
    int getMinimumDifference(TreeNode* root) {
        vec.clear();
        traversal(root);
        if (vec.size() < 2) return 0;
        int result = INT_MAX;
        for (int i = 1; i < vec.size(); i++) { // 统计有序数组的最小差值
            result = min(result, vec[i] - vec[i-1]);
        }
        return result;
    }
};

//双指针法
/**
 * 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 result = INT_MAX;
    TreeNode *pre = NULL;
    void traversal(TreeNode *root)
    {
        if(root == NULL)
        {
            return ;
        }
        traversal(root->left);
        if(pre != NULL)
        {
            result = min(result,(root->val - pre->val));
        }
        pre = root;
        traversal(root->right);
    }
    int getMinimumDifference(TreeNode* root) {
        traversal(root);
        return result;
    }
};

501.二叉搜索树中的众数

题目链接：https://leetcode.cn/problems/find-mode-in-binary-search-tree/description/

文章讲解：https://programmercarl.com/0501.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E4%B8%AD%E7%9A%84%E4%BC%97%E6%95%B0.html

视频讲解：https://www.bilibili.com/video/BV1fD4y117gp/

class Solution {
private:
    int maxCount = 0; // 最大频率
    int count = 0; // 统计频率
    TreeNode* pre = NULL;
    vector<int> result;
    void searchBST(TreeNode* cur) {
        if (cur == NULL) return ;

        searchBST(cur->left);       // 左
                                    // 中
        if (pre == NULL) { // 第一个节点
            count = 1;
        } else if (pre->val == cur->val) { // 与前一个节点数值相同
            count++;
        } else { // 与前一个节点数值不同
            count = 1;
        }
        pre = cur; // 更新上一个节点

        if (count == maxCount) { // 如果和最大值相同，放进result中
            result.push_back(cur->val);
        }

        if (count > maxCount) { // 如果计数大于最大值频率
            maxCount = count;   // 更新最大频率
            result.clear();     // 很关键的一步，不要忘记清空result，之前result里的元素都失效了
            result.push_back(cur->val);
        }

        searchBST(cur->right);      // 右
        return ;
    }

public:
    vector<int> findMode(TreeNode* root) {
        count = 0;
        maxCount = 0;
        pre = NULL; // 记录前一个节点
        result.clear();

        searchBST(root);
        return result;
    }
};

236. 二叉树的最近公共祖先

题目链接：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/description/

文章讲解：https://programmercarl.com/0236.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88.html

视频讲解：https://www.bilibili.com/video/BV1jd4y1B7E2/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == NULL)
        {
            return root;
        }
        if(root == q || root == p)
        {
            return root;
        }
        
        TreeNode *left = lowestCommonAncestor(root->left,p,q);
        TreeNode *right = lowestCommonAncestor(root->right,p,q);
        
        if(left != NULL && right != NULL)
        {
            return root;
        }
        else if(left != NULL && right == NULL)
        {
            return left;
        }
        else if(left == NULL && right != NULL)
        {
            return right;
        }
        else{
            return NULL;
        }
    }
};