LeetCode 솔루션 분류
669. Trim a Binary Search Tree
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[ LeetCode 시즌 3 ] 2022년 4월 14일 문제입니다.
[Medium] 669. Trim a Binary Search Tree
Given the root
of a binary search tree and the lowest and highest boundaries as low
and high
, trim the tree so that all its elements lies in [low, high]
. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.
Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.
Example 1:
Input: root = [1,0,2], low = 1, high = 2 Output: [1,null,2]
Example 2:
Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3 Output: [3,2,null,1]
Constraints:
- The number of nodes in the tree in the range
[1, 104]
. 0 <= Node.val <= 104
- The value of each node in the tree is unique.
root
is guaranteed to be a valid binary search tree.0 <= low <= high <= 104
관련자료
-
링크
댓글 4
Coffee님의 댓글
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재귀적 접근
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
if(root == null){
return null;
}
if(root.val > high){
return trimBST(root.left, low, high);
}
if(root.val < low){
return trimBST(root.right, low, high);
}
root.left = trimBST(root.left, low, high);
root.right = trimBST(root.right, low, high);
return root;
}
}
9dea0936님의 댓글
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class Solution:
def trimBST(self, root: Optional[TreeNode], low: int, high: int) -> Optional[TreeNode]:
def rec(root):
if not root:
return None
if root.val < low:
return rec(root.right)
elif root.val > high:
return rec(root.left)
else:
root.left = rec(root.left)
root.right = rec(root.right)
return root
return rec(root)
mingki님의 댓글
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C++
Runtime: 19 ms, faster than 66.98% of C++ online submissions for Trim a Binary Search Tree.
Memory Usage: 24 MB, less than 7.08% of C++ online submissions for Trim a Binary Search Tree.
Runtime: 19 ms, faster than 66.98% of C++ online submissions for Trim a Binary Search Tree.
Memory Usage: 24 MB, less than 7.08% of C++ online submissions for Trim a Binary Search Tree.
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int low, int high) {
if (!root) {
return root;
}
if (root->val < low) {
return trimBST(root->right, low, high);
}
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;
}
};
bobkim님의 댓글
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Runtime: 20 ms, faster than 46.21% of C++ online submissions for Trim a Binary Search Tree.
Memory Usage: 17.3 MB, less than 94.24% of C++ online submissions for Trim a Binary Search Tree.
Memory Usage: 17.3 MB, less than 94.24% of C++ online submissions for Trim a Binary Search Tree.
class Solution {
public:
void leftTrim(TreeNode* &p,const int &low, const int &high){
if(p->left != nullptr)
leftTrim(p->left,low,high);
if(p->right != nullptr)
rightTrim(p->right,low,high);
if(p->val < low){
p=p->right;
}else if(p->val > high){
p=p->left;
};
return ;
};
void rightTrim(TreeNode* &p, const int &low, const int &high){
if(p->left != nullptr)
leftTrim(p->left,low,high);
if(p->right != nullptr)
rightTrim(p->right,low,high);
if(p->val > high){
p=p->left;
}else if(p->val < low){
p=p->right;
};
return ;
};
TreeNode* trimBST(TreeNode* root, int low, int high) {
if(root->val < low){
leftTrim(root,low,high);
}else{
rightTrim(root,low,high);
};
return root;
}
};