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 depth of the two subtrees of every node never differ by more than 1.
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
Return true.
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
明白了平衡二叉树的定义,则对于每个结点,都需要判断其左右子树的高度,因此需要借助一个求高度的函数。
# 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):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
left = self.maxheight(root.left) # 当前结点左子树最大深度
right = self.maxheight(root.right) # 当前结点右子树最大深度
if abs(left - right) <= 1: # 如果当前结点满足平衡二叉树的条件
return self.isBalanced(root.left) and self.isBalanced(root.right) # 当前结点的左右结点也必须满足平衡二叉树的条件
else:
return False
def maxheight(self, root): # 找到左右子树最大深度
if root == None:
return 0
return max(self.maxheight(root.left), self.maxheight(root.right)) + 1