110. Balanced Binary Tree

Easy

Problem:

Given a binary tree, determine if it is .

Input: root = [3,9,20,null,null,15,7]
Output: true

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false
Explanation: The left subtree, 2, of 1(root) and the right subtree, 2, have a height difference of 2. Therefore, it returns false.

https://leetcode.com/problems/balanced-binary-tree/

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:
    result = True
    def isBalanced(self, root: Optional[TreeNode]) -> bool:
        def dfs(node):
            if node is None:
                return 0

            left = dfs(node.left)
            right = dfs(node.right)

            if abs(left - right) > 1:
                self.result = False
                return self.result
            
            return max(left, right) + 1

        if root:
            dfs(root)
            
        # Input: root = [], Output: true
        return self.result

A solution without using global variable result

# 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 isBalanced(self, root: Optional[TreeNode]) -> bool:
        def dfs(node):
            if node is None:
                return 0

            left = dfs(node.left)
            right = dfs(node.right)

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

        return dfs(root) != -1