LeetCode 1382: Balance a Binary Search Tree Solution

Master LeetCode problem 1382 (Balance a Binary Search Tree), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1382. Balance a Binary Search Tree

Medium

Divide and Conquer Greedy Tree Depth-First Search Binary Search Tree Binary Tree

Problem Explanation

Explanation

To balance a binary search tree, we can perform an in-order traversal of the original BST to collect all the nodes into a sorted array. Then, we can recursively build a balanced BST from the sorted array by selecting the middle element as the root and recursively constructing the left and right subtrees. This approach ensures the balanced property of the resulting binary search tree.

Algorithm:

Perform an in-order traversal of the original BST to collect all nodes into a sorted array.
Recursively build a balanced BST from the sorted array:
- Select the middle element as the root.
- Recursively build the left subtree from the elements to the left of the middle.
- Recursively build the right subtree from the elements to the right of the middle.

Time Complexity: O(n) where n is the number of nodes in the tree.

Space Complexity: O(n) for storing the nodes in the sorted array.

Solution Code

class Solution {
    public TreeNode balanceBST(TreeNode root) {
        List<TreeNode> nodes = new ArrayList<>();
        inOrderTraversal(root, nodes);
        return buildBalancedBST(nodes, 0, nodes.size() - 1);
    }
    
    private void inOrderTraversal(TreeNode node, List<TreeNode> nodes) {
        if (node == null) return;
        inOrderTraversal(node.left, nodes);
        nodes.add(node);
        inOrderTraversal(node.right, nodes);
    }
    
    private TreeNode buildBalancedBST(List<TreeNode> nodes, int start, int end) {
        if (start > end) return null;
        
        int mid = start + (end - start) / 2;
        TreeNode root = nodes.get(mid);
        
        root.left = buildBalancedBST(nodes, start, mid - 1);
        root.right = buildBalancedBST(nodes, mid + 1, end);
        
        return root;
    }
}

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