From de7c67c35a5d3d0e423072d0b7cb02876c628786 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Wed, 3 May 2023 20:13:38 -0500
Subject: [PATCH] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89=E6=A0=91?=
 =?UTF-8?q?=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?=
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---
 problems/0102.二叉树的层序遍历.md | 33 ++++++++++++-----------
 1 file changed, 18 insertions(+), 15 deletions(-)

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 13694207..c2ad9508 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
@@ -2938,23 +2938,26 @@ Python 3：
 #         self.right = right
 class Solution:
     def minDepth(self, root: TreeNode) -> int:
-        if root == None:
+        if not root:
             return 0
+        depth = 0
+        queue = collections.deque([root])
+        
+        while queue:
+            depth += 1 
+            for _ in range(len(queue)):
+                node = queue.popleft()
+                
+                if not node.left and not node.right:
+                    return depth
+            
+                if node.left:
+                    queue.append(node.left)
+                    
+                if node.right:
+                    queue.append(node.right)
 
-        #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
+        return depth
 ```
 
 Go：