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<int>& 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<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  

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;}
};