Update 0108.将有序数组转换为二叉搜索树.md

补充python注释

problems/0108.将有序数组转换为二叉搜索树.md

@@ -304,22 +304,42 @@ class Solution {
 }
 ```
 
-## Python
+## Python  
+**递归**
 
-递归法：
 ```python3
+# 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 sortedArrayToBST(self, nums: List[int]) -> TreeNode:
-        def buildaTree(left,right):
+        '''
-            if left > right: return None  #左闭右闭的区间，当区间 left > right的时候，就是空节点,当left = right的时候，不为空
+        构造二叉树：重点是选取数组最中间元素为分割点，左侧是递归左区间；右侧是递归右区间
-            mid = left + (right - left) // 2 #保证数据不会越界
+        必然是平衡树
-            val = nums[mid]
+        左闭右闭区间
-            root = TreeNode(val)
+        '''
-            root.left = buildaTree(left,mid - 1)
+        # 返回根节点
-            root.right = buildaTree(mid + 1,right)
+        root = self.traversal(nums, 0, len(nums)-1)
-            return root
-        root = buildaTree(0,len(nums) - 1)  #左闭右闭区间
         return root
+
+    def traversal(self, nums: List[int], left: int, right: int) -> TreeNode:
+        # Base Case
+        if left > right:
+            return None
+        
+        # 确定左右界的中心，防越界
+        mid = left + (right - left) // 2
+        # 构建根节点
+        mid_root = TreeNode(nums[mid])
+        # 构建以左右界的中心为分割点的左右子树
+        mid_root.left = self.traversal(nums, left, mid-1)
+        mid_root.right = self.traversal(nums, mid+1, right)
+
+        # 返回由被传入的左右界定义的某子树的根节点
+        return mid_root
 ```
 
 ## Go