feat: add python and java solution to lcof problem

添加《剑指 Offer》题解：面试题28. 对称的二叉树

Author: yanglbme

lcof/面试题28. 对称的二叉树/README.md

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+# [面试题28. 对称的二叉树](https://leetcode-cn.com/problems/dui-cheng-de-er-cha-shu-lcof/)
+
+## 题目描述
+请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样，那么它是对称的。
+
+例如，二叉树 `[1,2,2,3,4,4,3]` 是对称的。
+
+```
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+```
+
+但是下面这个 `[1,2,2,null,3,null,3]` 则不是镜像对称的:
+
+```
+    1
+   / \
+  2   2
+   \   \
+   3    3
+```
+
+**示例 1：**
+
+```
+输入：root = [1,2,2,3,4,4,3]
+输出：true
+```
+
+**示例 2：**
+
+```
+输入：root = [1,2,2,null,3,null,3]
+输出：false
+```
+
+**限制：**
+
+- `0 <= 节点个数 <= 1000`
+
+## 解法
+### Python3
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if root is None:
+            return True
+        return self.symmetric(root.left, root.right)
+    
+    def symmetric(self, node1, node2) -> bool:
+        if node1 is None and node2 is None:
+            return True
+        if node1 is None or node2 is None or node1.val != node2.val:
+            return False
+        return self.symmetric(node1.left, node2.right) and self.symmetric(node1.right, node2.left)
+```
+
+### Java
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isSymmetric(TreeNode root) {
+        if (root == null) {
+            return true;
+        }
+        return isSymmetric(root.left, root.right);
+    }
+
+    private boolean isSymmetric(TreeNode node1, TreeNode node2) {
+        if (node1 == null && node2 == null) {
+            return true;
+        }
+        if (node1 == null || node2 == null || node1.val != node2.val) {
+            return false;
+        }
+        return isSymmetric(node1.left, node2.right) && isSymmetric(node1.right, node2.left);
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题28. 对称的二叉树/Solution.java

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+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isSymmetric(TreeNode root) {
+        if (root == null) {
+            return true;
+        }
+        return isSymmetric(root.left, root.right);
+    }
+
+    private boolean isSymmetric(TreeNode node1, TreeNode node2) {
+        if (node1 == null && node2 == null) {
+            return true;
+        }
+        if (node1 == null || node2 == null || node1.val != node2.val) {
+            return false;
+        }
+        return isSymmetric(node1.left, node2.right) && isSymmetric(node1.right, node2.left);
+    }
+}

lcof/面试题28. 对称的二叉树/Solution.py

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+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if root is None:
+            return True
+        return self.symmetric(root.left, root.right)
+    
+    def symmetric(self, node1, node2) -> bool:
+        if node1 is None and node2 is None:
+            return True
+        if node1 is None or node2 is None or node1.val != node2.val:
+            return False
+        return self.symmetric(node1.left, node2.right) and self.symmetric(node1.right, node2.left)