# 1382. Balance-a-Binary-Search-Tree

difficulty: Medium

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Given a binary search tree, return a **balanced** binary search tree with the same node values.

A binary search tree is *balanced* if and only if the depth of the two subtrees of every node never differ by more than 1.

If there is more than one answer, return any of them.

**Example 1:**

![](https://assets.leetcode.com/uploads/2019/08/22/1515_ex1.png)![](https://assets.leetcode.com/uploads/2019/08/22/1515_ex1_out.png)

```
Input: root = [1,null,2,null,3,null,4,null,null]
Output: [2,1,3,null,null,null,4]
Explanation: This is not the only correct answer, [3,1,4,null,2,null,null] is also correct.
```

**Constraints:**

* The number of nodes in the tree is between `1` and `10^4`.
* The tree nodes will have distinct values between `1` and `10^5`.

## Method One

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private List<TreeNode> inorder;
    public TreeNode balanceBST(TreeNode root) {
        this.inorder = new ArrayList<>();
        inorderTraversal(root);
        return balanceBST(0, inorder.size() - 1);
    }
    public void inorderTraversal(TreeNode node){
        if(node == null){
            return;
        }
        inorderTraversal(node.left);
        inorder.add(node);
        inorderTraversal(node.right);
    }
    public TreeNode balanceBST(int start, int end){
        if( start > end ){
            return null;
        }
        int pivot = start + (end - start)/2;
        TreeNode node = inorder.get(pivot);
        node.left = balanceBST(start,pivot -1 );
        node.right = balanceBST(pivot + 1, end);
        return node;
    }
}
​
```
