# 1382. Balance a Binary Search Tree

## Description

Given the root of a binary search tree, return a balanced binary search tree with the same node values. If there is more than one answer, return any of them.

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

Example 1:

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] is also correct.


Example 2:

Input: root = [2,1,3]
Output: [2,1,3]


Constraints:

• The number of nodes in the tree is in the range [1, 104].
• 1 <= Node.val <= 105

## Solutions

Solution 1: In-order Traversal + Construct Balanced Binary Search Tree

 Since the original tree is a binary search tree, we can save the result of the in-order traversal in an array $nums$. Then we design a function $build(i, j)$, which is used to construct a balanced binary search tree within the index range $[i, j]$ in $nums$. The answer is $build(0, nums - 1)$.

The execution logic of the function $build(i, j)$ is as follows:

• If $i > j$, then the balanced binary search tree is empty, return an empty node;
• Otherwise, we take $mid = (i + j) / 2$ as the root node, then recursively build the left and right subtrees, and return the root node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary search tree.

• /**
* 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<Integer> nums = new ArrayList<>();

public TreeNode balanceBST(TreeNode root) {
dfs(root);
return build(0, nums.size() - 1);
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
dfs(root.right);
}

private TreeNode build(int i, int j) {
if (i > j) {
return null;
}
int mid = (i + j) >> 1;
TreeNode left = build(i, mid - 1);
TreeNode right = build(mid + 1, j);
return new TreeNode(nums.get(mid), left, right);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* balanceBST(TreeNode* root) {
dfs(root);
return build(0, nums.size() - 1);
}

private:
vector<int> nums;

void dfs(TreeNode* root) {
if (!root) {
return;
}
dfs(root->left);
nums.push_back(root->val);
dfs(root->right);
}

TreeNode* build(int i, int j) {
if (i > j) {
return nullptr;
}
int mid = (i + j) >> 1;
TreeNode* left = build(i, mid - 1);
TreeNode* right = build(mid + 1, j);
return new TreeNode(nums[mid], left, right);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def balanceBST(self, root: TreeNode) -> TreeNode:
def dfs(root: TreeNode):
if root is None:
return
dfs(root.left)
nums.append(root.val)
dfs(root.right)

def build(i: int, j: int) -> TreeNode:
if i > j:
return None
mid = (i + j) >> 1
left = build(i, mid - 1)
right = build(mid + 1, j)
return TreeNode(nums[mid], left, right)

nums = []
dfs(root)
return build(0, len(nums) - 1)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func balanceBST(root *TreeNode) *TreeNode {
ans := []int{}
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
ans = append(ans, root.Val)
dfs(root.Right)
}
var build func(i, j int) *TreeNode
build = func(i, j int) *TreeNode {
if i > j {
return nil
}
mid := (i + j) >> 1
left := build(i, mid-1)
right := build(mid+1, j)
return &TreeNode{Val: ans[mid], Left: left, Right: right}
}
dfs(root)
return build(0, len(ans)-1)
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function balanceBST(root: TreeNode | null): TreeNode | null {
const nums: number[] = [];
const dfs = (root: TreeNode | null): void => {
if (root == null) {
return;
}
dfs(root.left);
nums.push(root.val);
dfs(root.right);
};
const build = (i: number, j: number): TreeNode | null => {
if (i > j) {
return null;
}
const mid: number = (i + j) >> 1;
const left: TreeNode | null = build(i, mid - 1);
const right: TreeNode | null = build(mid + 1, j);
return new TreeNode(nums[mid], left, right);
};
dfs(root);
return build(0, nums.length - 1);
}