Trim a Binary Search Tree
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.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
| 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
|
java Solution:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode trimBST(TreeNode root, int L, int R) {
if(root == null){
return root;
}
if(root.val > R){
return trimBST(root.left,L,R);
}
else if(root.val < L){
return trimBST(root.right,L,R);
}
else{
root.left = trimBST(root.left,L,R);
root.right = trimBST(root.right,L,R);
}
return root;
}
}
|
Kotlin Solution:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
| fun trimBST(root: TreeNode?, L: Int, R: Int): TreeNode? {
if (root == null) {
return root
}
if (root.`val` > R) {
return trimBST(root.left, L, R)
} else if (root.`val` < L) {
return trimBST(root.right, L, R)
} else {
root.left = trimBST(root.left, L, R)
root.right = trimBST(root.right, L, R)
}
return root
}
|
