From 4b71311621baad2dc356dbe100e608435cc47bd8 Mon Sep 17 00:00:00 2001
From: shuwen <shuwenlle@foxmail.com>
Date: Thu, 19 Aug 2021 12:41:16 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E5=B1=82=E5=BA=8F?=
 =?UTF-8?q?=E9=81=8D=E5=8E=86=E4=B8=AD=E7=9A=84=20104.=20=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6?=
 =?UTF-8?q?=20Python=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0102.二叉树的层序遍历.md | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)

diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 0f9b2df6..dc386684 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -1532,6 +1532,29 @@ Java：
 
 
 Python：
+```python 3
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        if root == None:
+            return 0
+        
+        queue_ = [root]
+        result = []
+        while queue_:
+            length = len(queue_)
+            sub = []
+            for i in range(length):
+                cur = queue_.pop(0)
+                sub.append(cur.val)
+                #子节点入队列
+                if cur.left: queue_.append(cur.left)
+                if cur.right: queue_.append(cur.right)
+            result.append(sub)
+            
+
+        return len(result)
+```
+
 
 Go：
 

From 0990dfe810af83c4ed41c605fd165dad1f76645b Mon Sep 17 00:00:00 2001
From: shuwen <shuwenlle@foxmail.com>
Date: Thu, 19 Aug 2021 17:00:42 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200102.=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?=
 =?UTF-8?q?.md=20=20=E5=85=B3=E4=BA=8E111.=E4=BA=8C=E5=8F=89=E6=A0=91?=
 =?UTF-8?q?=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6=E7=9A=84Python?=
 =?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0102.二叉树的层序遍历.md | 32 ++++++++++++++++++++++-
 1 file changed, 31 insertions(+), 1 deletion(-)

diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index dc386684..109ed70f 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -1562,6 +1562,8 @@ JavaScript：
 
 # 111.二叉树的最小深度
 
+题目地址：https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/
+
 相对于 104.二叉树的最大深度 ，本题还也可以使用层序遍历的方式来解决，思路是一样的。
 
 **需要注意的是，只有当左右孩子都为空的时候，才说明遍历的最低点了。如果其中一个孩子为空则不是最低点**
@@ -1597,7 +1599,35 @@ public:
 Java： 
 
 
-Python：
+Python 3：
+
+```python 3
+# 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 minDepth(self, root: TreeNode) -> int:
+        if root == None:
+            return 0
+
+        #根节点的深度为1
+        queue_ = [(root,1)]
+        while queue_:
+            cur, depth = queue_.pop(0)
+            
+            if cur.left == None and cur.right == None:
+                return depth
+            #先左子节点，由于左子节点没有孩子，则就是这一层了
+            if cur.left:
+                queue_.append((cur.left,depth + 1))
+            if cur.right:
+                queue_.append((cur.right,depth + 1))
+
+        return 0
+```
 
 Go：