Balance a Binary Search Tree

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

SOLUTION:

# 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 inorder(self, root):
        if root:
            self.inorder(root.left)
            self.tree.append(root.val)
            self.inorder(root.right)

    def arrToTree(self, arr, i, j):
        if i <= j - 1:
            root = TreeNode()
            mid = (i + j) // 2
            root.val = arr[mid]
            root.left = self.arrToTree(arr, i, mid)
            root.right = self.arrToTree(arr, mid + 1, j)
            return root
        return None

    def balanceBST(self, root: TreeNode) -> TreeNode:
        self.tree = []
        self.inorder(root)
        return self.arrToTree(self.tree, 0, len(self.tree))

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Balance a Binary Search Tree

Abhishek Chaudhary

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