[LeetCode] 669. Trim a Binary Search Tree
grandyang opened this issue · comments
Given a binary search tree and the lowest and highest boundaries as L
and R
, trim the tree so that all its elements lies in [L, R]
(R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.
Example 1:
Input:
1
/ \
0 2
L = 1
R = 2
Output:
1
\
2
Example 2:
Input:
3
/ \
0 4
\
2
/
1
L = 1
R = 3
Output:
3
/
2
/
1
这道题让我们修剪一棵二叉搜索树,给了个边界范围[L, R], 所有不在这个范围内的结点应该被移除掉,但是仍需要保留二叉搜索树的性质,即左<根<右,有时候是小于等于。博主最开始的想法是先遍历一遍二叉树,将在返回内的结点值都放到一个数组后,遍历结束后再根据数组重建一棵二叉搜索树。这种方法会在某些test case上fail掉,可能会改变原来的二叉搜索树的结构,所以我们只能换一种思路。正确方法其实应该是在遍历的过程中就修改二叉树,移除不合题意的结点。当然对于二叉树的题,十有八九都是要用递归来解的。首先判断如果root为空,那么直接返回空即可。然后就是要看根结点是否在范围内,如果根结点值小于L,那么返回对其右子结点调用递归函数的值;如果根结点大于R,那么返回对其左子结点调用递归函数的值。如果根结点在范围内,将其左子结点更新为对其左子结点调用递归函数的返回值,同样,将其右子结点更新为对其右子结点调用递归函数的返回值。最后返回root即可,参见代码如下:
解法一:
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int L, int R) {
if (!root) return NULL;
if (root->val < L) return trimBST(root->right, L, R);
if (root->val > R) return trimBST(root->left, L, R);
root->left = trimBST(root->left, L, R);
root->right = trimBST(root->right, L, R);
return root;
}
};
下面这种方法是迭代的写法,虽然树的题一般都是用递归来写,简洁又美观。但是我们也可以强行用while来代替递归,比如下面这种写法:
解法二:
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int L, int R) {
if (!root) return NULL;
while (root->val < L || root->val > R) {
root = (root->val < L) ? root->right : root->left;
}
TreeNode *cur = root;
while (cur) {
while (cur->left && cur->left->val < L) {
cur->left = cur->left->right;
}
cur = cur->left;
}
cur = root;
while (cur) {
while (cur->right && cur->right->val > R) {
cur->right = cur->right->left;
}
cur = cur->right;
}
return root;
}
};
参考资料:
https://discuss.leetcode.com/topic/102034/java-solution-6-liner
https://discuss.leetcode.com/topic/104140/java-solution-iteration-version