Balance a Binary Search Tree — LC Medium

1382. 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

# 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:
        # traverse the tree and store it to the list
        if not root:
            return None
        
        tempList = []
        
        def helperDFS(node):
            
            if not node:
                return
            
            helperDFS(node.left)
            tempList.append(node.val)
            helperDFS(node.right)
        
        helperDFS(root)
        
        # once you have printed all the nodes in sorted order, start from the median node and start attaching it
        
        return self.BSTbuilder(tempList)
    
    def BSTbuilder(self, tempLists):
        if not tempLists:
            return
        mid = len(tempLists) // 2
        roots = TreeNode(tempLists[mid])
        roots.left = self.BSTbuilder(tempLists[:mid])
        roots.right = self.BSTbuilder(tempLists[mid + 1:])
        
        return roots