feat: add python and java solutions to lcci question

添加《程序员面试金典》题解：面试题 04.04. 检查平衡性

Author: yanglbme

lcci/04.04.Check Balance/README.md

@@ -6,20 +6,61 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+递归法。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```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 not root:
+            return True
+        l, r = self._height(root.left), self._height(root.right)
+        return abs(l - r) < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)
 
+    def _height(self, node):
+        if not node:
+            return 0
+        return 1 + max(self._height(node.left), self._height(node.right))
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```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;
+        }
+        int l = height(root.left), r = height(root.right);
+        return Math.abs(l - r) < 2 && isBalanced(root.left) && isBalanced(root.right);
+    }
 
+    private int height(TreeNode node) {
+        if (node == null) {
+            return 0;
+        }
+        return 1 + Math.max(height(node.left), height(node.right));
+    }
+}
 ```
 
 ### ...

lcci/04.04.Check Balance/README_EN.md

@@ -1,53 +1,122 @@
-# [04.04. Check Balance](https://leetcode-cn.com/problems/check-balance-lcci)
-
-## Description
-<p>Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.</p>
-
-<p><br />
-<strong>Example 1:</strong></p>
-
-<pre>
-Given tree [3,9,20,null,null,15,7]
-    3
-   / \
-  9  20
-    /  \
-   15   7
-return true.</pre>
-
-<p><strong>Example 2:</strong></p>
-
-<pre>
-Given [1,2,2,3,3,null,null,4,4]
-      1
-     / \
-    2   2
-   / \
-  3   3
- / \
-4   4
-return&nbsp;false.</pre>
-
-<p>&nbsp;</p>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [04.04. Check Balance](https://leetcode-cn.com/problems/check-balance-lcci)
+
+## Description
+<p>Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.</p>
+
+
+
+<p><br />
+
+<strong>Example 1:</strong></p>
+
+
+
+<pre>
+
+Given tree [3,9,20,null,null,15,7]
+
+    3
+
+   / \
+
+  9  20
+
+    /  \
+
+   15   7
+
+return true.</pre>
+
+
+
+<p><strong>Example 2:</strong></p>
+
+
+
+<pre>
+
+Given [1,2,2,3,3,null,null,4,4]
+
+      1
+
+     / \
+
+    2   2
+
+   / \
+
+  3   3
+
+ / \
+
+4   4
+
+return&nbsp;false.</pre>
+
+
+
+<p>&nbsp;</p>
+
+
+
+
+## Solutions
+
+
+### 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 not root:
+            return True
+        l, r = self._height(root.left), self._height(root.right)
+        return abs(l - r) < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)
+
+    def _height(self, node):
+        if not node:
+            return 0
+        return 1 + max(self._height(node.left), self._height(node.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;
+        }
+        int l = height(root.left), r = height(root.right);
+        return Math.abs(l - r) < 2 && isBalanced(root.left) && isBalanced(root.right);
+    }
+
+    private int height(TreeNode node) {
+        if (node == null) {
+            return 0;
+        }
+        return 1 + Math.max(height(node.left), height(node.right));
+    }
+}
+```
+
+### ...
+```
+
+```

lcci/04.04.Check Balance/Solution.java

@@ -0,0 +1,25 @@
+/**
+ * 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;
+        }
+        int l = height(root.left), r = height(root.right);
+        return Math.abs(l - r) < 2 && isBalanced(root.left) && isBalanced(root.right);
+    }
+
+    private int height(TreeNode node) {
+        if (node == null) {
+            return 0;
+        }
+        return 1 + Math.max(height(node.left), height(node.right));
+    }
+}

lcci/04.04.Check Balance/Solution.py

@@ -0,0 +1,19 @@
+# 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 not root:
+            return True
+        l, r = self._height(root.left), self._height(root.right)
+        return abs(l - r) < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)
+
+    def _height(self, node):
+        if not node:
+            return 0
+        return 1 + max(self._height(node.left), self._height(node.right))