feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题55 - II. 平衡二叉树

Author: yanglbme

lcof/面试题55 - II. 平衡二叉树/README.md

@@ -0,0 +1,93 @@
+# [面试题55 - II. 平衡二叉树](https://leetcode-cn.com/problems/ping-heng-er-cha-shu-lcof/)
+
+## 题目描述
+输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过 1，那么它就是一棵平衡二叉树。
+
+**示例 1:**
+
+给定二叉树 `[3,9,20,null,null,15,7]`
+
+```
+    3
+   / \
+  9  20
+    /  \
+   15   7
+```
+
+返回 `true`。
+
+**示例 2:**
+
+给定二叉树 `[1,2,2,3,3,null,null,4,4]`
+
+```
+       1
+      / \
+     2   2
+    / \
+   3   3
+  / \
+ 4   4
+```
+
+返回 `false`。
+
+**限制：**
+
+- `1 <= 树的结点个数 <= 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 isBalanced(self, root: TreeNode) -> bool:
+        if root is None:
+            return True
+        return abs(self._height(root.left) - self._height(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)
+
+    def _height(self, tree):
+        if tree is None:
+            return 0
+        return 1 + max(self._height(tree.left), self._height(tree.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 isBalanced(TreeNode root) {
+        if (root == null) {
+            return true;
+        }
+        return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
+    }
+
+    private int height(TreeNode tree) {
+        if (tree == null) {
+            return 0;
+        }
+        return 1 + Math.max(height(tree.left), height(tree.right));
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题55 - II. 平衡二叉树/Solution.java

@@ -0,0 +1,24 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isBalanced(TreeNode root) {
+        if (root == null) {
+            return true;
+        }
+        return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
+    }
+
+    private int height(TreeNode tree) {
+        if (tree == null) {
+            return 0;
+        }
+        return 1 + Math.max(height(tree.left), height(tree.right));
+    }
+}

lcof/面试题55 - II. 平衡二叉树/Solution.py

@@ -0,0 +1,17 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def isBalanced(self, root: TreeNode) -> bool:
+        if root is None:
+            return True
+        return abs(self._height(root.left) - self._height(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)
+
+    def _height(self, tree):
+        if tree is None:
+            return 0
+        return 1 + max(self._height(tree.left), self._height(tree.right))