代码随想录算法训练营
- 代码随想录算法训练营43期 | Day
- 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
代码随想录算法训练营43期 | Day
108.将有序数组转换为二叉搜索树
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
class Solution {
public:
int pre = 0;
void traversal(TreeNode* cur)
{
// 终止条件
if(cur==nullptr) return;
//右
traversal(cur->right);
cur->val += pre;
pre = cur->val;
traversal(cur->left);
}
TreeNode* convertBST(TreeNode* root) {
pre = 0;
traversal(root);
return root;
}
};
