代码随想录算法训练营第二十一天|LeetCode669. 修剪二叉搜索树、LeetCode108.将有序数组转换为二叉搜索树、LeetCode538.把二叉搜索树转换为累加树-优快云博客

/**
 * 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:
    TreeNode* traversal(TreeNode* root, int low, int high) {
        if (root == nullptr) {
            return nullptr;
        }
        if (root->val < low) {
            TreeNode* right = traversal(root->right, low, high);
            return right;
        }
        if(root->val > high)
        {
            TreeNode* left = traversal(root->left, low, high);
            return left;
        }
        root->left = traversal(root->left, low, high);
        root->right = traversal(root->right, low, high);

        return root;
    }
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        return traversal(root, low, high);
    }
};

二、LeetCode108.将有序数组转换为二叉搜索树

题目链接：

108. 将有序数组转换为二叉搜索树 - 力扣（LeetCode）

代码：

/**
 * 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:
    TreeNode* traversal(vector<int>& nums, int left, int right)
    {
        if(left > right)
            return nullptr;
        
        int mid = (left + right) / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = traversal(nums, left, mid - 1);
        root->right = traversal(nums, mid + 1, right);
        return root;
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return traversal(nums, 0, nums.size() - 1);
    }
};

三、LeetCode538.把二叉搜索树转换为累加树

题目链接：

538. 把二叉搜索树转换为累加树 - 力扣（LeetCode）

代码：

/**
 * 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 curAcc = 0;
    // 整一个右根左遍历顺序
    void traversal(TreeNode* root)
    {
        if(root == nullptr) return ;

        traversal(root->right);
        root->val += curAcc;
        curAcc = root->val;
        traversal(root->left);
    }
    TreeNode* convertBST(TreeNode* root) {
        this->curAcc = 0;
        traversal(root);
        return root;
    }
};