669. 修剪二叉搜索树
其实就是判断当前节点的值与搜索区间的关系。
- 位于 [ l o w , h i g h ] [low, high] [low,high]: 继续遍历修剪左子树和右子树
- 位于 ( − ∞ , l o w ) (-\infty, low) (−∞,low): 返回修剪的右子树
- 位于 ( h i g h , + ∞ ) (high, +\infty) (high,+∞): 返回修剪的左子树
cpp
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int low, int high) {
if (root == nullptr)
return nullptr;
if (root->val < low)
return trimBST(root->right, low, high);
else if (root->val > high)
return trimBST(root->left, low, high);
else {
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
本质就是寻找分割点,分割点作为当前节点,然后递归左区间和右区间。
cpp
class Solution {
public:
TreeNode* construct(vector<int>& nums, int left, int right) {
//[left, right)
if (left >= right)
return nullptr;
int middle = (left + right)/2;
TreeNode* root = new TreeNode(nums[middle]);
//[left, middle)
root->left = construct(nums, left, middle);
//[middle + 1, right);
root->right = construct(nums, middle + 1, right);
return root;
}
TreeNode* sortedArrayToBST(vector<int>& nums) {
return construct(nums, 0, nums.size());
}
};
538.把二叉搜索树转换为累加树
cpp
class Solution {
public:
TreeNode* prev = nullptr;
TreeNode* convertBST(TreeNode* root) {
if (root == nullptr)
return nullptr;
root->right = convertBST(root->right);
if (prev != nullptr)
root->val += prev->val;
prev = root;
root->left = convertBST(root->left);
return root;
}
};