LeetCode #1382 — MEDIUM

Balance a Binary Search Tree

Move from brute-force thinking to an efficient approach using greedy strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy · Tree

Example 1

[1,null,2,null,3,null,4]

Example 2

[2,1,3]

Step 02

Core Insight

What unlocks the optimal approach

Convert the tree to a sorted array using an in-order traversal.
Construct a new balanced tree from the sorted array recursively.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1382: Balance a 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);
        nums.add(root.val);
        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);
    }
}

// Accepted solution for LeetCode #1382: Balance a Binary Search Tree
/**
 * 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)
}

# Accepted solution for LeetCode #1382: Balance a Binary Search Tree
# 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)

// Accepted solution for LeetCode #1382: Balance a Binary Search Tree
struct Solution;
use rustgym_util::*;

trait Inorder {
    fn inorder(self, nodes: &mut Vec<i32>);
    fn build(nodes: &[i32]) -> TreeLink;
}

impl Inorder for TreeLink {
    fn inorder(self, nodes: &mut Vec<i32>) {
        if let Some(node) = self {
            let mut node = node.borrow_mut();
            let val = node.val;
            let left = node.left.take();
            let right = node.right.take();
            left.inorder(nodes);
            nodes.push(val);
            right.inorder(nodes);
        }
    }
    fn build(nodes: &[i32]) -> TreeLink {
        let n = nodes.len();
        if n == 0 {
            None
        } else {
            let m = n / 2;
            tree!(
                nodes[m],
                TreeLink::build(&nodes[..m]),
                TreeLink::build(&nodes[m + 1..])
            )
        }
    }
}

impl Solution {
    fn balance_bst(root: TreeLink) -> TreeLink {
        let mut nodes: Vec<i32> = vec![];
        root.inorder(&mut nodes);
        TreeLink::build(&nodes)
    }
}

#[test]
fn test() {
    let root = tree!(1, None, tree!(2, None, tree!(3, None, tree!(4))));
    let res = tree!(3, tree!(2, tree!(1), None), tree!(4));
    assert_eq!(Solution::balance_bst(root), res);
}

// Accepted solution for LeetCode #1382: Balance a Binary Search Tree
/**
 * 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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.