add #26 #27 python implement

notes/面试总结.md

@@ -28,6 +28,8 @@
 * [23. 链表中环的入口结点](#23-链表中环的入口结点)
 * [24. 反转链表](#24-反转链表)
 * [25. 合并两个排序的链表](#25-合并两个排序的链表)
+* [26. 树的子结构](#26-树的子结构)
+* [27. 二叉树的镜像](#27-二叉树的镜像)
 
 * [参考文献](#参考文献)
 <!-- GFM-TOC -->
@@ -1965,6 +1967,108 @@ class Solution:
             cur.next = pHead2
         return head.next
 ```
+# 26. 树的子结构
+
+[NowCoder](https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+<div align="center"> <img src="../pics//4583e24f-424b-4d50-8a14-2c38a1827d4a.png" width="500"/> </div><br>
+
+## 解题思路
+
+```java
+public boolean HasSubtree(TreeNode root1, TreeNode root2) {
+    if (root1 == null || root2 == null)
+        return false;
+    return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
+}
+
+private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
+    if (root2 == null)
+        return true;
+    if (root1 == null)
+        return false;
+    if (root1.val != root2.val)
+        return false;
+    return isSubtreeWithRoot(root1.left, root2.left) && isSubtreeWithRoot(root1.right, root2.right);
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+class Solution:
+    def HasSubtree(self, pRoot1, pRoot2):
+        # write code here
+        if pRoot1 == None or pRoot2 == None:
+            return False
+        return self.isSubtreeWithRoot(pRoot1, pRoot2) or \
+               self.HasSubtree(pRoot1.left, pRoot2) or \
+               self.HasSubtree(pRoot1.right, pRoot2)
+    
+    def isSubtreeWithRoot(self, pRoot1, pRoot2):
+        if pRoot2 == None:
+            return True
+        if pRoot1 == None:
+            return False
+        if pRoot1.val != pRoot2.val:
+            return False
+        return self.isSubtreeWithRoot(pRoot1.left, pRoot2.left) and \
+               self.isSubtreeWithRoot(pRoot1.right, pRoot2.right)
+```
+
+# 27. 二叉树的镜像
+
+[NowCoder](https://www.nowcoder.com/practice/564f4c26aa584921bc75623e48ca3011?tpId=13&tqId=11171&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+<div align="center"> <img src="../pics//a2d13178-f1ef-4811-a240-1fe95b55b1eb.png" width="300"/> </div><br>
+
+## 解题思路
+
+```java
+public void Mirror(TreeNode root) {
+    if (root == null)
+        return;
+    swap(root);
+    Mirror(root.left);
+    Mirror(root.right);
+}
+
+private void swap(TreeNode root) {
+    TreeNode t = root.left;
+    root.left = root.right;
+    root.right = t;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+class Solution:
+    # 返回镜像树的根节点
+    def Mirror(self, root):
+        # write code here
+        if root == None:
+            return
+        self.swap(root)
+        self.Mirror(root.left)
+        self.Mirror(root.right)
+        
+        
+    def swap(self, root):
+        root.left, root.right = root.right, root.left
+```
 
 # 参考文献