add #28 #29 python implement

notes/面试总结.md

@@ -30,6 +30,8 @@
 * [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)
+* [28 对称的二叉树](#28-对称的二叉树)
+* [29. 顺时针打印矩阵](#29-顺时针打印矩阵)
 
 * [参考文献](#参考文献)
 <!-- GFM-TOC -->
@@ -2069,6 +2071,119 @@ class Solution:
     def swap(self, root):
         root.left, root.right = root.right, root.left
 ```
+# 28 对称的二叉树
+
+[NowCder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+<div align="center"> <img src="../pics//f42443e0-208d-41ea-be44-c7fd97d2e3bf.png" width="300"/> </div><br>
+
+## 解题思路
+
+```java
+boolean isSymmetrical(TreeNode pRoot) {
+    if (pRoot == null)
+        return true;
+    return isSymmetrical(pRoot.left, pRoot.right);
+}
+
+boolean isSymmetrical(TreeNode t1, TreeNode t2) {
+    if (t1 == null && t2 == null)
+        return true;
+    if (t1 == null || t2 == null)
+        return false;
+    if (t1.val != t2.val)
+        return false;
+    return isSymmetrical(t1.left, t2.right) && isSymmetrical(t1.right, t2.left);
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+class Solution:
+    def isSymmetrical(self, pRoot):
+        # write code here
+        if pRoot == None:
+            return True
+        return self._isSymmetrical(pRoot.left, pRoot.right)
+    
+    def _isSymmetrical(self, t1, t2):
+        if t1 == None and t2 == None:
+            return True
+        if t1 == None or t2 == None:
+            return False
+        if t1.val != t2.val:
+            return False
+        return self._isSymmetrical(t1.left, t2.right) and self._isSymmetrical(t1.right, t2.left)
+```
+
+# 29. 顺时针打印矩阵
+
+[NowCoder](https://www.nowcoder.com/practice/9b4c81a02cd34f76be2659fa0d54342a?tpId=13&tqId=11172&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
+
+## 题目描述
+
+下图的矩阵顺时针打印结果为：1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
+
+<div align="center"> <img src="../pics//6539b9a4-2b24-4d10-8c94-2eb5aba1e296.png" width="300"/> </div><br>
+
+## 解题思路
+
+```java
+public ArrayList<Integer> printMatrix(int[][] matrix) {
+    ArrayList<Integer> ret = new ArrayList<>();
+    int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
+    while (r1 <= r2 && c1 <= c2) {
+        for (int i = c1; i <= c2; i++)
+            ret.add(matrix[r1][i]);
+        for (int i = r1 + 1; i <= r2; i++)
+            ret.add(matrix[i][c2]);
+        if (r1 != r2)
+            for (int i = c2 - 1; i >= c1; i--)
+                ret.add(matrix[r2][i]);
+        if (c1 != c2)
+            for (int i = r2 - 1; i > r1; i--)
+                ret.add(matrix[i][c1]);
+        r1++; r2--; c1++; c2--;
+    }
+    return ret;
+}
+```
+
+```python
+# -*- coding:utf-8 -*-
+class Solution:
+    # matrix类型为二维列表，需要返回列表
+    def printMatrix(self, matrix):
+        # write code here
+        len_row = len(matrix)
+        len_column = len(matrix[0])
+        print_list = []
+        if(len_row == 0):
+            return print_list
+        for i in range(0,(min(len_row,len_column)-1)/2+1):#圈数
+            #四条边打印+终止检测
+            #注意打印个数处理即可
+            for j in range(i,len_column-i):
+                print_list.append(matrix[i][j])
+            for j in range(i+1,len_row-i):
+                print_list.append(matrix[j][len_column-i-1])
+            if (len_row-1-i == i):
+                return print_list
+            for j in range(i+1,len_column-i):
+                print_list.append(matrix[len_row-i-1][len_column-j-1])
+            if (len_column-1-i == i):
+                return print_list
+            for j in range(i+1,len_row-i-1):
+                print_list.append(matrix[len_row-j-1][i])
+        return print_list
+```
 
 # 参考文献