代码随想录二刷 Day23

669. 修剪二叉搜索树

找到小数字的右子树与大数字左子树必须要重新检查一遍然后让root的左右直接指向return的左右节点；

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == NULL) return NULL;
        if (root->val < low) {
           // root->left = trimBST(root->right, low, high);
           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. Convert Sorted Array to Binary Search Tree

这个题尝试把 root->left = traversal(nums, left, mid - 1); 里面的left改成0但是不行，因为left和right的数值在迭代的过程中会被mid动态替换；

class Solution {
public:

    // 左闭右闭区间[left, right]
    TreeNode* traversal(vector& nums, int left, int right) {
        if (left > right) return NULL; //这句话是因为下面mid可能减一或者加一，然后那mid减完之后还可能变成left或者right，所下面的left和right不能改成0和nums.size() -1;
        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;
    }

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

538.把二叉搜索树转换为累加树  

注意要右中左来遍历二叉树， 中节点的处理逻辑就是让cur的数值加上前一个节点的数值，就是反中序来做累加；

class Solution {
private:
    int pre = 0; // 记录前一个节点的数值
    void traversal(TreeNode* cur) { // 右中左遍历
        if (cur == NULL) return;
        traversal(cur->right);
        cur->val += pre;
        pre = cur->val;
        traversal(cur->left);
    }
public:
    TreeNode* convertBST(TreeNode* root) {
        traversal(root);
        return root;
    }
};