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+# Algorithm Homework 9
+
+> 姓名：阙建明
+>
+> 学号：1901213139
+
+## 作业要求：
+
+Leetcode: 235, 110, 257
+
+## 题解：
+
+- ### Leetcode problem 235
+
+  >  [二叉搜索树的最近公共祖先](https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/)
+
+  - **算法思路：** 
+
+    1. 首先判断节点p和节点q的大小，小的记作left，大的记作right；
+    2. 接着对二叉搜索进行递归前序遍历；
+    3. 根据二叉搜索树的特性，目标节点的值要满足大于等于left节点的值，且小于等于right节点的值
+
+  - **Leetcode提交结果：**
+
+    ![image-20200507221022918](AlgorithmHomework9.assets/image-20200507221022918.png)
+
+  - **代码：**
+
+    ```python
+    class Solution(object):
+        def lowestCommonAncestor(self, root, p, q):
+            """
+            :type root: TreeNode
+            :type p: TreeNode
+            :type q: TreeNode
+            :rtype: TreeNode
+            """
+    
+            def preOrderTraversal(root, left, right):
+                if root is None or (left.val <= root.val <= right.val):
+                    return root
+    
+                leftResult = preOrderTraversal(root.left, left, right)
+                rightResult = preOrderTraversal(root.right, left, right)
+                return leftResult if leftResult is not None else rightResult
+    
+            left = p if p.val < q.val else q
+            right = p if p.val >= q.val else q
+            return preOrderTraversal(root, left, right)
+    ```
+
+- ### Leetcode problem 110
+
+  > [平衡二叉树](https://leetcode-cn.com/problems/balanced-binary-tree/)
+
+  - **算法思路：** 
+
+    1. 直接递归计算每个节点左右子树的高度；
+    2. 根据AVL树的特性，如果任意一个节点的左右子树的高度差大于一，则该树不是AVL树，标记器高度为-1;
+    3. 如果某个子树的高度计算为-1,则表示该子树不是AVL树，-1会一直传递到顶层。
+
+  - **Leetcode提交结果：**
+
+  ![image-20200507222442113](AlgorithmHomework9.assets/image-20200507222442113.png)
+
+  - **代码：**
+
+    ```python
+    class Solution(object):
+        def isBalanced(self, root):
+            """
+            :type root: TreeNode
+            :rtype: bool
+            """
+            def preOrderTraversal(root):
+                if root is None:
+                    return 0
+                leftNum = preOrderTraversal(root.left)
+                rightNum = preOrderTraversal(root.right)
+                if leftNum == -1 or rightNum == -1 or abs(leftNum - rightNum) > 1:
+                    return -1
+                return 1 + max(leftNum, rightNum)
+    
+            return preOrderTraversal(root) != -1
+    ```
+
+- ### Leetcode problem 729
+
+  > [二叉树的所有路径](https://leetcode-cn.com/problems/binary-tree-paths/)
+
+  - **算法思路：** 
+
+    1. 直接使用递归进行深度搜索；
+  2. 一旦检索到一个叶子节点，就将当前遍历的路径添加到结果当中。
+  
+- **Leetcode提交结果：**
+  
+  ![image-20200507225110291](AlgorithmHomework9.assets/image-20200507225110291.png)
+  
+- **代码：**
+  
+    ```python
+    class Solution(object):
+        def binaryTreePaths(self, root):
+            """
+            :type root: TreeNode
+            :rtype: List[str]
+            """
+    
+            def preOrderTraversal(root, s, list):
+                if root is None:
+                    return
+                if root.left is None and root.right is None:
+                    list.append(s + "->" + str(root.val))
+                preOrderTraversal(root.left, s + "->" + str(root.val), list)
+                preOrderTraversal(root.right, s + "->" + str(root.val), list)
+    
+            if root is None:
+                return []
+            if root.left is None and root.right is None:
+                return [str(root.val)]
+            s = str(root.val)
+            list = []
+                
+            preOrderTraversal(root.left, s, list)
+            preOrderTraversal(root.right, s, list)
+            return list
+    ```
+    
+    
\ No newline at end of file
diff --git a/README.md b/README.md
index ba7b450..2ca4e39 100755
--- a/README.md
+++ b/README.md
@@ -8,4 +8,5 @@
 - [第六次作业](AlgorithmHomework6.md)
 - [第七次作业](AlgorithmHomework7.md)
 - [第八次作业](AlgorithmHomework8.md)
+- [第九次作业](AlgorithmHomework9.md)