2024-06-26 15:39:14 +0000 UTC
Balance a Binary Search Tree
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# 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:
self.sortedArr = []
self.inorderTraverse(root)
return self.sortedArrayToBST(0, len(self.sortedArr) - 1)
def inorderTraverse(self, root: TreeNode) -> None:
if not root:
return
self.inorderTraverse(root.left)
self.sortedArr.append(root)
self.inorderTraverse(root.right)
def sortedArrayToBST(self, start: int, end: int) -> TreeNode:
if start > end:
return None
mid = (start + end) // 2
root = self.sortedArr[mid]
root.left = self.sortedArrayToBST(start, mid - 1)
root.right = self.sortedArrayToBST(mid + 1, end)
return root