QUESTION:
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
EXPLANATION:
自己写的办法比较蠢啦,哈哈,可以看看别人的算法,确实还是挺快。速度上倒是没有什么差距,但是明显的递归思想的应用上,确实还是有点欠缺。
SOLUTION:
class Solution {
TreeNode trimResult = null;
public TreeNode trimBST(TreeNode root, int L, int R) {
if(root == null) return trimResult;
if(root.val>=L && root.val<=R) trimResult = new TreeNode(root.val);
if(root.left!=null) TrimBSTHelper(root.left,L,R,trimResult);
if(root.right!=null) TrimBSTHelper(root.right,L,R,trimResult);
return trimResult;
}
public void TrimBSTHelper(TreeNode root,int L,int R,TreeNode result){
if(root == null) return;
if(root.val>=L && root.val <=R) {
if (result == null){
trimResult = new TreeNode(root.val);
if(root.left!=null) TrimBSTHelper(root.left,L,R,trimResult);
if(root.right!=null) TrimBSTHelper(root.right,L,R,trimResult);
}else{
if (root.val < result.val) {
result.left = new TreeNode(root.val);
if (root.left != null) TrimBSTHelper(root.left, L, R, result.left);
if (root.right != null) TrimBSTHelper(root.right, L, R, result.left);
}
if (root.val > result.val) {
result.right = new TreeNode(root.val);
if (root.left != null) TrimBSTHelper(root.left, L, R, result.right);
if (root.right != null) TrimBSTHelper(root.right, L, R, result.right);
}
}
}else{
if (root.left != null) TrimBSTHelper(root.left, L, R, result);
if (root.right != null) TrimBSTHelper(root.right, L, R, result);
}
}
}
class Solution {
public TreeNode trimBST(TreeNode root, int L, int R) {
if (root == null) 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;
}
}