题目要求
思路一:模拟迭代
- 依次判断每个节点是否合法:
- 首先找出结果的根,若原根小了就拉右边的过来,大了拉左边的过来做新根;
- 然后分别判断左右子树的大小,由于二叉搜索树的性质,子树只需要判断一边就好:
- 左子树判断是否>low,合法就向左下走,不合法往右下;
- 右子树判断是否<high,合法就向右下走,不合法往左下。
Java
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
while (root != null && (root.val < low || root.val > high)) // 确定原根是否合法
root = root.val < low ? root.right : root.left;
TreeNode res = root;
while (root != null) {
while (root.left != null && root.left.val < low)
root.left = root.left.right;
root = root.left;
}
root = res;
while (root != null) {
while (root.right != null && root.right.val > high)
root.right = root.right.left;
root = root.right;
}
return res;
}
}
- 时间复杂度:O(n)
- 空间复杂度:O(1)
C++
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int low, int high) {
while (root != nullptr && (root->val < low || root->val > high)) // 确定原根是否合法
root = root->val < low ? root->right : root->left;
TreeNode* res = root;
while (root != nullptr) {
while (root->left != nullptr && root->left->val < low)
root->left = root->left->right;
root = root->left;
}
root = res;
while (root != nullptr) {
while (root->right != nullptr && root->right->val > high)
root->right = root->right->left;
root = root->right;
}
return res;
}
};
- 时间复杂度:O(n)
- 空间复杂度:O(1)
思路二:递归
- 直接用当前函数递归修剪即可:
- 当前值小了放右下(大)的值进去,剪掉当前和左边节点;
- 当前值大了放左下(小)的值进去,剪掉当前和右边节点。
- 然后递归掉下面所有节点。
Java
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
if (root == null)
return null;
if (root.val < low)
return trimBST(root.right, low, high);
else if (root.val > high)
return trimBST(root.left, low, high);
root.left = trimBST(root.left, low, high);
root.right = trimBST(root.right, low, high);
return root;
}
}
- 时间复杂度:O(n)
- 空间复杂度:O(1),忽略递归的额外空间开销
C++
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);
root->left = trimBST(root->left, low, high);
root->right = trimBST(root->right, low, high);
return root;
}
};
- 时间复杂度:O(n)
- 空间复杂度:O(1),忽略递归的额外空间开销
Rust
- 今天又见识到了新报错:
already borrowed: BorrowMutError
,不能把borrow
的东西来回随便等,要搞临时中间变量,闭包要关好。
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn trim_bst(root: Option<Rc<RefCell<TreeNode>>>, low: i32, high: i32) -> Option<Rc<RefCell<TreeNode>>> {
if root.is_none() {
return None;
}
if root.as_ref().unwrap().borrow().val < low {
return Solution::trim_bst(root.as_ref().unwrap().borrow().right.clone(), low, high);
}
else if root.as_ref().unwrap().borrow().val > high {
return Solution::trim_bst(root.as_ref().unwrap().borrow().left.clone(), low, high);
}
let (l, r) = (Solution::trim_bst(root.as_ref().unwrap().borrow().left.clone(), low, high), Solution::trim_bst(root.as_ref().unwrap().borrow().right.clone(), low, high)); // 要先拎出来
root.as_ref().unwrap().borrow_mut().left = l;
root.as_ref().unwrap().borrow_mut().right = r;
root
}
}
- 时间复杂度:O(n)
- 空间复杂度:O(1),忽略递归的额外空间开销
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