From 3d59f732e0479c12657f5a6020b4ed5784222780 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 01:35:13 -0500 Subject: [PATCH] =?UTF-8?q?Update=200222.=E5=AE=8C=E5=85=A8=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E8=8A=82=E7=82=B9=E4=B8=AA=E6=95=B0?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0222.完全二叉树的节点个数.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/problems/0222.完全二叉树的节点个数.md b/problems/0222.完全二叉树的节点个数.md index d89a9bce..795a6f37 100644 --- a/problems/0222.完全二叉树的节点个数.md +++ b/problems/0222.完全二叉树的节点个数.md @@ -393,6 +393,20 @@ class Solution: # 利用完全二叉树特性 return 2**count-1 return 1+self.countNodes(root.left)+self.countNodes(root.right) ``` +完全二叉树写法3 +```python +class Solution: # 利用完全二叉树特性 + def countNodes(self, root: TreeNode) -> int: + if not root: return 0 + count = 0 + left = root.left; right = root.right + while left and right: + count+=1 + left = left.left; right = right.right + if not left and not right: # 如果同时到底说明是满二叉树，反之则不是 + return (2<