题解 | #判断是不是平衡二叉树#

思路：

1、判断当前节点是否平衡：左右子树高度之差小于等于1

2、递归判断全部节点是否平衡

class Solution:
    def IsBalanced_Solution(self , pRoot: TreeNode) -> bool:
        
        return self.check(pRoot)
        
    def check(self, p):
        if not p:
            return True
        
        left = self.getHigh(p.left)
        right = self.getHigh(p.right)
        
        return abs(left - right) <= 1 and self.check(p.left) and self.check(p.right)
        
    def getHigh(self, p):
        if not p:
            return 0
        
        left = self.getHigh(p.left)
        right = self.getHigh(p.right)
        
        return max(left, right) + 1