110. 平衡二叉树
给定一个二叉树，判断它是否是高度平衡的二叉树。
本题中，一棵高度平衡二叉树定义为：
一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。
示例 1：
输入：root = [3,9,20,null,null,15,7]
输出：true
示例 2：
输入：root = [1,2,2,3,3,null,null,4,4]
输出：false
示例 3：
输入：root = []
输出：true
提示：
树中的节点数在范围 [0, 5000] 内
-104 <= Node.val <= 104
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/balanced-binary-tree/
Link：https://leetcode.com/problems/balanced-binary-tree/
递归
O(N)
应该先判断左右子树是否是平衡的，然后还要记录左右子树的高度，这样才能判断当前节点是否平衡
# 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: TreeNode) -> bool:
return self.helper(root)[0]
def helper(self, node:TreeNode) -> Tuple[bool, int]:
if node is None:
return (True, 0)
left_balanced, left_height = self.helper(node.left)
right_balanced, right_height = self.helper(node.right)
height = max(left_height, right_height) + 1
if left_balanced and right_balanced and abs(left_height - right_height) < 2:
return (True, height)
else:
return (False, height)
另一种, 使用负数-1代表不平衡, 整数代表平衡
# 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: TreeNode) -> bool:
return self.helper(root) != -1
def helper(self, node:TreeNode) -> int:
if node is None:
return 0
left_height = self.helper(node.left)
right_height = self.helper(node.right)
if left_height != -1 and right_height != -1 and abs(left_height - right_height) < 2:
return max(left_height, right_height) + 1
else:
return -1
–End–