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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