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.
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.
LeetCode:链接
按照平衡二叉树的定义,有两点需要注意:1,每个节点的两个子树也都是平衡二叉树;2,每个节点的两个子树的高度差不超过1。 直接的思路是:遍历每个节点,判断每个节点的两个子树高度差是否小于等于1(第一个递归);而求一个子树的高度可以用递归法求解(第二个递归)。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isBalanced(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
if abs(self.getDepth(root.left) - self.getDepth(root.right)) > 1:
return False
return self.isBalanced(root.left) and self.isBalanced(root.right)
def getDepth(self, root):
if root == None:
return 0
return 1 + max(self.getDepth(root.left), self.getDepth(root.right))