【刷题】代码随想录算法训练营第二十三天|669、修剪二叉搜索树,108、将有序数组转换为二叉搜索树,538、把二叉搜索树转换为累加树

发布于:2024-04-28 ⋅ 阅读:(16) ⋅ 点赞:(0)

669、修剪二叉搜索树

讲解:https://programmercarl.com/0669.%E4%BF%AE%E5%89%AA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html]

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class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == NULL) return root;
        if(root->val < low){
            TreeNode* right = trimBST(root->right, low, high);
            return right;
        }
        if (root->val > high){
            TreeNode* left = trimBST(root->left, low, high);
            return left;
        }
        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);

        return root;
    }
};

108、将有序数组转换为二叉搜索树

讲解:https://programmercarl.com/0108.%E5%B0%86%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84%E8%BD%AC%E6%8D%A2%E4%B8%BA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html

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


public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode* root = traversal(nums, 0, nums.size()-1);
        return root;
    }
};

538、把二叉搜索树转换为累加树

讲解:https://programmercarl.com/0538.%E6%8A%8A%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E8%BD%AC%E6%8D%A2%E4%B8%BA%E7%B4%AF%E5%8A%A0%E6%A0%91.html

题目示例:
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class Solution {
private:
    int pre = 0;
    void traversal(TreeNode* cur){
        if(cur == NULL) return;
        traversal(cur->right);
        cur->val += pre;
        pre = cur->val;
        traversal(cur->left);
    }

public:
    TreeNode* convertBST(TreeNode* root) {
        pre = 0;
        traversal(root);
        return root;
    }
};