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# 669.Trim-a-Binary-Search-Tree

## 题目描述

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

## 代码

### Approach #1 Recursion

Time Complexity:O(N) && Space Complexity: O(N)
/**
* 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);
if (root.val < L) return trimBST(root.right, L, R);
root.left = trimBST(root.left, L, R);
root.right = trimBST(root.right, L, R);
return root;
}
}

### Approach #2 Iteration

class Solution {
public TreeNode trimBST(TreeNode root, int L, int R) {
if (root == null) {
return root;
}
// 1. Find a valid root which is used to return.
while (root.val < L || root.val > R) {
if (root.val < L) {
root = root.right;
}
if (root.val > R) {
root = root.left;
}
}
TreeNode dummy = root;
// 2. Remove the invalid nodes from left subtree.
while (dummy != null) {
while (dummy.left != null && dummy.left.val < L) {
dummy.left = dummy.left.right;
// If the left child is smaller than L, then we just keep the right subtree of it.
}
dummy = dummy.left;
}
dummy = root;
// 3. Remove the invalid nodes from right subtree
while (dummy != null) {
while (dummy.right != null && dummy.right.val > R) {
dummy.right = dummy.right.left;
// If the right child is biggrt than R, then we just keep the left subtree of it.
}
dummy = dummy.right;
}
return root;
}
}