题目1:669. 修剪二叉搜索树 - 力扣(LeetCode)
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int low, int high) {
if(root == NULL) return NULL;
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;
}
};
题目2:108. 将有序数组转换为二叉搜索树 - 力扣(LeetCode)
class Solution {
public:
TreeNode* traversal(vector<int>& nums, int left, int right) {
if(left > right) return NULL;
int mid = (left + right) / 2;
TreeNode* node = new TreeNode(nums[mid]);
node->left = traversal(nums, left, mid - 1);
node->right = traversal(nums, mid + 1, right);
return node;
}
TreeNode* sortedArrayToBST(vector<int>& nums) {
return traversal(nums, 0, nums.size() - 1);
}
};
题目3:538. 把二叉搜索树转换为累加树 - 力扣(LeetCode)
class Solution {
public:
int pre = 0;
void traversal(TreeNode* node) {
if(node == NULL) return;
traversal(node->right);
node->val = node->val + pre;
pre = node->val;
traversal(node->left);
}
TreeNode* convertBST(TreeNode* root) {
traversal(root);
return root;
}
};