Description
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the left and right subtrees of every node differ in height by no more than 1.
Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
Solutions
1. 递归
# Definition for a binary tree node.
# Time: O(n)
# Space: O(1)
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
is_flag, height = self._is_balanced(root)
return True if is_flag else False
def _is_balanced(self, root):
# return is_balanced, height
if not root:
return True, 0
left_b, left_h = self._is_balanced(root.left)
right_b, right_h = self._is_balanced(root.right)
if not left_b or not right_b:
return False, -1
else:
if abs(left_h - right_h) > 1:
return False, -1
else:
return True, max(left_h, right_h) + 1
# 227/227 cases passed (36 ms)
# Your runtime beats 99.66 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (16.4 MB)
2. 递归-优化
# Time Complexity: O(n)
# Space Complexity: O(n)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
if not root:
return True
self.balanced = True
self.dfs_height(root)
return self.balanced
def dfs_height(self, root):
if not root or not self.balanced:
return -1
l = self.dfs_height(root.left)
r = self.dfs_height(root.right)
if abs(l-r) > 1:
self.balanced = False
return -1
return max(l, r) + 1
# Runtime: 56 ms, faster than 87.77% of Python3 online submissions for Balanced Binary Tree.
# Memory Usage: 18.6 MB, less than 45.71% of Python3 online submissions for Balanced Binary Tree.