目录
110. 平衡二叉树
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解题卡点:
# 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: Optional[TreeNode]) -> bool:
def get_height(node):
if node is None:
return 0
left_node = get_height(node.left)
if left_node == -1:
return -1
right_node = get_height(node.right)
if right_node == -1:
return -1
if abs(left_node - right_node) > 1:
return -1
else:
return 1 + max(left_node, right_node)
if get_height(root) == -1:
return False
else:
return True
257. 二叉树的所有路径
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404. 左叶子之和
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222. 完全二叉树的节点个数
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