Balanced Binary Tree
Problem
https://leetcode.com/problems/balanced-binary-tree/
Given a binary tree, determine if it is height-balanced.
Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: true
Example 2:

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false
Example 3:
Input: root = []
Output: true
Constraints:
The number of nodes in the tree is in the range
[0, 5000].-10:sup:`4`<= Node.val <= 10:sup:`4`
Pattern
Tree, Depth-First Search, Binary Tree
Approaches
Code
from __future__ import annotations
class TreeNode:
"""Node in a binary tree."""
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
@classmethod
def from_list(cls, vals: list[int | None]) -> TreeNode | None:
if not vals:
return None
root = TreeNode(vals[0])
queue = [root]
i = 1
while i < len(vals):
node = queue.pop(0)
if i < len(vals) and vals[i] is not None:
node.left = TreeNode(vals[i])
queue.append(node.left)
i += 1
if i < len(vals) and vals[i] is not None:
node.right = TreeNode(vals[i])
queue.append(node.right)
i += 1
return root
def isBalanced(root: TreeNode | None) -> bool:
"""Return whether the binary tree is height-balanced."""
def dfs(node: TreeNode | None) -> int:
if node is None:
return 0
left = dfs(node.left)
if left == -1:
return -1
right = dfs(node.right)
if right == -1:
return -1
if abs(left - right) > 1:
return -1
return 1 + max(left, right)
return dfs(root) != -1
Test
>>> from balanced_binary_tree__dfs import TreeNode, isBalanced
>>> isBalanced(TreeNode.from_list([3, 9, 20, None, None, 15, 7]))
True
>>> isBalanced(TreeNode.from_list([1, 2, 2, 3, 3, None, None, 4, 4]))
False
>>> isBalanced(TreeNode.from_list([]))
True
- class balanced_binary_tree__dfs.TreeNode(val=0, left=None, right=None)
Bases:
objectNode in a binary tree.