From b4f55e72745f4f6fbf27dd4bdb2201f1b430b9f2 Mon Sep 17 00:00:00 2001
From: zhangchang <zhangchang@stu.cqie.edu.cn>
Date: Sat, 24 Feb 2024 22:46:14 +0800
Subject: [PATCH] =?UTF-8?q?=E4=BC=98=E5=8C=96Python=E9=80=92=E5=BD=92?=
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---
 problems/二叉树的递归遍历.md | 54 +++++++++++++++++-----------
 1 file changed, 33 insertions(+), 21 deletions(-)

diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md
index edd55aad..4621d4a7 100644
--- a/problems/二叉树的递归遍历.md
+++ b/problems/二叉树的递归遍历.md
@@ -180,26 +180,34 @@ class Solution {
 
 class Solution:
     def preorderTraversal(self, root: TreeNode) -> List[int]:
-        if not root:
-            return []
-
-        left = self.preorderTraversal(root.left)
-        right = self.preorderTraversal(root.right)
-
-        return  [root.val] + left +  right
+        res = []
+        
+        def dfs(node):
+            if node is None:
+                return
+            
+            res.append(node.val)
+            dfs(node.left)
+            dfs(node.right)
+        
+        return res
 
 ```
 ```python
 # 中序遍历-递归-LC94_二叉树的中序遍历
 class Solution:
     def inorderTraversal(self, root: TreeNode) -> List[int]:
-        if root is None:
-            return []
-
-        left = self.inorderTraversal(root.left)
-        right = self.inorderTraversal(root.right)
-
-        return left + [root.val] + right
+        res = []
+        
+        def dfs(node):
+            if node is None:
+                return
+            
+            dfs(node.left)
+            res.append(node.val)
+            dfs(node.right)
+        
+        return res
 ```
 ```python
 
@@ -207,13 +215,17 @@ class Solution:
 # 后序遍历-递归-LC145_二叉树的后序遍历
 class Solution:
     def postorderTraversal(self, root: TreeNode) -> List[int]:
-        if not root:
-            return []
-
-        left = self.postorderTraversal(root.left)
-        right = self.postorderTraversal(root.right)
-
-        return left + right + [root.val]
+        res = []
+        
+        def dfs(node):
+            if node is None:
+                return
+            
+            dfs(node.left)
+            dfs(node.right)
+            res.append(node.val)
+        
+        return res
 ```
 
 ### Go：