feat: add python and java solution to lcof problem

添加《剑指 Offer》题解：面试题26. 树的子结构

Author: yanglbme

lcof/面试题26. 树的子结构/README.md

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+# [面试题26. 树的子结构](https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof/)
+
+## 题目描述
+输入两棵二叉树 A 和 B，判断 B 是不是 A W的子结构。(约定空树不是任意一个树的子结构)
+
+B 是 A 的子结构， 即 A 中有出现和 B 相同的结构和节点值。
+
+**例如:**
+
+给定的树 A:
+
+```
+     3
+    / \
+   4   5
+  / \
+ 1   2
+```
+
+给定的树 B：
+
+```
+   4 
+  /
+ 1
+```
+
+返回 true，因为 B 与 A 的一个子树拥有相同的结构和节点值。
+
+**示例 1：**
+
+```
+输入：A = [1,2,3], B = [3,1]
+输出：false
+```
+
+**示例 2：**
+
+```
+输入：A = [3,4,5,1,2], B = [4,1]
+输出：true
+```
+
+**限制：**
+
+- `0 <= 节点个数 <= 10000`
+
+## 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 isSubStructure(self, A: TreeNode, B: TreeNode) -> bool:
+        return self.sub(A, B) if B else False
+
+    def sub(self, A: TreeNode, B: TreeNode) -> bool:
+        if B is None:
+            return True
+        if A is None:
+            return False
+        if A.val == B.val:
+            return self.same(A, B) or self.sub(A.left, B) or self.sub(A.right, B)
+        return self.sub(A.left, B) or self.sub(A.right, B)
+
+    def same(self, A: TreeNode, B: TreeNode) -> bool:
+        if B is None:
+            return True
+        if A is None or A.val != B.val:
+            return False
+        return self.same(A.left, B.left) and self.same(A.right, B.right)
+```
+
+### 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 isSubStructure(TreeNode A, TreeNode B) {
+        return B == null ? false : sub(A, B);
+    }
+
+    private boolean sub(TreeNode A, TreeNode B) {
+        if (B == null) {
+            return true;
+        }
+        if (A == null) {
+            return false;
+        }
+        if (A.val == B.val) {
+            return isSame(A, B) || sub(A.left, B) || sub(A.right, B);
+        }
+        return sub(A.left, B) || sub(A.right, B);
+        
+    }
+
+    private boolean isSame(TreeNode A, TreeNode B) {
+        if (B == null) {
+            return true;
+        }
+        if (A == null || A.val != B.val) {
+            return false;
+        }
+        return isSame(A.left, B.left) && isSame(A.right, B.right);
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题26. 树的子结构/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 isSubStructure(TreeNode A, TreeNode B) {
+        return B == null ? false : sub(A, B);
+    }
+
+    private boolean sub(TreeNode A, TreeNode B) {
+        if (B == null) {
+            return true;
+        }
+        if (A == null) {
+            return false;
+        }
+        if (A.val == B.val) {
+            return isSame(A, B) || sub(A.left, B) || sub(A.right, B);
+        }
+        return sub(A.left, B) || sub(A.right, B);
+        
+    }
+
+    private boolean isSame(TreeNode A, TreeNode B) {
+        if (B == null) {
+            return true;
+        }
+        if (A == null || A.val != B.val) {
+            return false;
+        }
+        return isSame(A.left, B.left) && isSame(A.right, B.right);
+    }
+}

lcof/面试题26. 树的子结构/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 isSubStructure(self, A: TreeNode, B: TreeNode) -> bool:
+        return self.sub(A, B) if B else False
+
+    def sub(self, A: TreeNode, B: TreeNode) -> bool:
+        if B is None:
+            return True
+        if A is None:
+            return False
+        if A.val == B.val:
+            return self.same(A, B) or self.sub(A.left, B) or self.sub(A.right, B)
+        return self.sub(A.left, B) or self.sub(A.right, B)
+
+    def same(self, A: TreeNode, B: TreeNode) -> bool:
+        if B is None:
+            return True
+        if A is None or A.val != B.val:
+            return False
+        return self.same(A.left, B.left) and self.same(A.right, B.right)