110. Balanced Binary Tree

1 min readApr 6, 2022


checkout the problem description here.


Recursive approaches

Approach 1:

Naive approach

Iterate through the tree and find left height and right height of the subtree(another helper function that calculates the height is called) and check if balanced. Check for its left and right node and so on.

A kind of top down approach.

if root
return true
left_height = height_find(node.left)+1
right_height =height_find(node.right)+1
if abs(left_height -right_height) >1
return false
left = check(node.left)
right = check(node.right)

if left ==false or right == false
return false
return true

Time : O(n²)

Space : O(n) at max if skew tree

Approach 2:

The more optimized way

Recursively iterate and find the height of left and right subtree and if the height differ by more than 1 , return a -1.Which is a flag used to indicate that the tree is unbalanced.

A kind of bottom up approach.

class Solution:
def isBalanced(self, root: Optional[TreeNode]) -> bool:
if self.rec(root) ==-1:
return False
return True

def rec(self,root):
if not root:
return 0
left = self.rec(root.left)

if left==-1 or right == -1 or abs(right-left)>1 :
return -1
return 1 + max(left,right)

Time : O(n)

Space : O(n) at max if skew tree




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