以下题目均来自LeetCode。
输入一棵二叉树的根节点,判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1,那么它就是一棵平衡二叉树。
示例 1:
给定二叉树 [3,9,20,null,null,15,7]
3
/ \
9 20
/ \
15 7
Answer:
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
def recur(root):
if not root: return 0
left = recur(root.left)
if left == -1: return -1
right = recur(root.right)
if right == -1: return -1
return max(left, right) + 1 if abs(left - right) <= 1 else -1
return recur(root) != -1
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