Update 110._balanced_binary_tree.md

Author: 0xkookoo

docs/Leetcode_Solutions/110._balanced_binary_tree.md

@@ -1,37 +1,68 @@
-###110. Balanced Binary Tree
-
-题目:
-<https://leetcode.com/problems/balanced-binary-tree/>
-
-
-难度:
-Easy
-
-
-全程递归中
-
-
-
-```
+#  110. Balanced Binary Tree
+**<font color=red>难度: 简单</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/balanced-binary-tree/description/
+
+> 内容描述
+
+```
+Given a binary tree, determine if it is height-balanced.
+
+For this problem, a height-balanced binary tree is defined as:
+
+a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+Example 1:
+
+Given the following tree [3,9,20,null,null,15,7]:
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+Return true.
+
+Example 2:
+
+Given the following tree [1,2,2,3,3,null,null,4,4]:
+
+       1
+      / \
+     2   2
+    / \
+   3   3
+  / \
+ 4   4
+Return false.
+```
+
+## 解题方案
+
+> 思路 1
+
+递归,判断左右子树最大高度差不超过1且左右子树均为平衡树
+
+```python
 class Solution(object):
     def isBalanced(self, root):
         """
         :type root: TreeNode
         :rtype: bool
         """
-        if root == None:
+        def height(node):
+            if not node:
+                return 0
+            return 1 + max(height(node.left), height(node.right))
+        if not root:
             return True
-        
-        lh = self.height(root.left)
-        rh = self.height(root.right)
-        
-        if abs(lh-rh) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right):
-            return True
-        return False
-        
-        
-    def height(self, node):
-        if node == None:
-            return 0
-        return 1 + max(self.height(node.left),self.height(node.right))
-```
+        return abs(height(root.left) - height(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)
+```
+
+
+
+