3
LeetCode: 110. Balanced Binary Tree
source link: https://mozillazg.com/2021/01/leetcode-110-balanced-binary-tree.html
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
题目¶
原题地址：https://leetcode.com/problems/balanced-binary-tree/
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:
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].
- -104 <= Node.val <= 104
解法¶
求左右子树各自的最大深度，然后进行比较，如果它们的深度差小于等于1的话， 则当前层是平衡的，继续比较下一层。
这个方法的 Python 代码类似下面这样:
# 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):
if root is None:
return True
left_height = self.max_height(root.left)
right_height = self.max_height(root.right)
return abs(left_height - right_height) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)
def max_height(self, root):
if root is None:
return 0
return 1 + max(self.max_height(root.left), self.max_height(root.right))
Recommend
About Joyk
Aggregate valuable and interesting links.
Joyk means Joy of geeK