From db18da9914e231bd3063215b61353cd931f8ffa2 Mon Sep 17 00:00:00 2001
From: ironartisan <ironerhalo@gmail.com>
Date: Wed, 11 Aug 2021 22:24:28 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A01382.=E5=B0=86=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E5=8F=98=E5=B9=B3=E8=A1=A1?=
 =?UTF-8?q?python3=E4=BB=A3=E7=A0=81?=
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---
 problems/1382.将二叉搜索树变平衡.md | 24 ++++++++++++++++++--
 1 file changed, 22 insertions(+), 2 deletions(-)

diff --git a/problems/1382.将二叉搜索树变平衡.md b/problems/1382.将二叉搜索树变平衡.md
index b9d4fb65..75530663 100644
--- a/problems/1382.将二叉搜索树变平衡.md
+++ b/problems/1382.将二叉搜索树变平衡.md
@@ -55,7 +55,7 @@ private:
         vec.push_back(cur->val);
         traversal(cur->right);
     }
-    有序数组转平衡二叉树
+    // 有序数组转平衡二叉树
     TreeNode* getTree(vector<int>& nums, int left, int right) {
         if (left > right) return nullptr;
         int mid = left + ((right - left) / 2);
@@ -78,7 +78,27 @@ public:
 Java：
 
 Python：
-
+```python
+class Solution:
+    def balanceBST(self, root: TreeNode) -> TreeNode:
+        res = []
+        # 有序树转成有序数组
+        def traversal(cur: TreeNode):
+            if not cur: return
+            traversal(cur.left)
+            res.append(cur.val)
+            traversal(cur.right)
+        # 有序数组转成平衡二叉树
+        def getTree(nums: List, left, right):
+            if left > right: return 
+            mid = left + (right -left) // 2
+            root = TreeNode(nums[mid])
+            root.left = getTree(nums, left, mid - 1)
+            root.right = getTree(nums, mid + 1, right)
+            return root
+        traversal(root)
+        return getTree(res, 0, len(res) - 1)
+```
 Go：
 
 JavaScript：