feat: add solutions to lc problem: No.0863,1704

* No.0863.All Nodes Distance K in Binary Tree
* No.1704.Determine if String Halves Are Alike

Author: yanglbme

solution/0800-0899/0860.Lemonade Change/README.md

@@ -53,6 +53,12 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：贪心 + 模拟**
+
+从前往后遍历账单数组 `bills`，如果当前账单是 5 美元，那么直接收下即可；如果当前账单是 10 美元，那么需要找零 5 美元；如果当前账单是 20 美元，那么需要找零 15 美元，此时有两种找零方式：找零 1 张 10 美元 + 1 张 5 美元；找零 3 张 5 美元。如果找零失败，直接返回 `false`。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为账单数组 `bills` 的长度。
+
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 ### **Python3**

solution/0800-0899/0863.All Nodes Distance K in Binary Tree/README.md

@@ -50,7 +50,11 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-先用哈希表存放每个节点的父节点，然后 DFS 找到距离目标节点 target 为 k 的节点即可。
+**方法一：DFS + 哈希表**
+
+我们先用 DFS 遍历整棵树，记录每个结点的父结点，然后从目标结点开始，向上、向下分别搜索距离为 $k$ 的结点，添加到答案数组中。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的结点数。
 
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@@ -97,6 +101,40 @@ class Solution:
         return ans
 ```
 
+```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 distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
+        def dfs1(root, fa):
+            if root is None:
+                return
+            p[root] = fa
+            dfs1(root.left, root)
+            dfs1(root.right, root)
+
+        def dfs2(root, fa, k):
+            if root is None:
+                return
+            if k == 0:
+                ans.append(root.val)
+                return
+            for nxt in (root.left, root.right, p[root]):
+                if nxt != fa:
+                    dfs2(nxt, root, k - 1)
+
+        p = {}
+        dfs1(root, None)
+        ans = []
+        dfs2(target, None, k)
+        return ans
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->

solution/0800-0899/0863.All Nodes Distance K in Binary Tree/README_EN.md

@@ -80,6 +80,40 @@ class Solution:
         return ans
 ```
 
+```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 distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
+        def dfs1(root, fa):
+            if root is None:
+                return
+            p[root] = fa
+            dfs1(root.left, root)
+            dfs1(root.right, root)
+        
+        def dfs2(root, fa, k):
+            if root is None:
+                return
+            if k == 0:
+                ans.append(root.val)
+                return
+            for nxt in (root.left, root.right, p[root]):
+                if nxt != fa:
+                    dfs2(nxt, root, k - 1)
+
+        p = {}
+        dfs1(root, None)
+        ans = []
+        dfs2(target, None, k)
+        return ans
+```
+
 ### **Java**
 
 ```java

solution/1700-1799/1704.Determine if String Halves Are Alike/README.md

@@ -68,6 +68,14 @@ class Solution:
         return cnt == 0
 ```
 
+```python
+class Solution:
+    def halvesAreAlike(self, s: str) -> bool:
+        vowels = set('aeiouAEIOU')
+        a, b = s[:len(s) >> 1], s[len(s) >> 1:]
+        return sum(c in vowels for c in a) == sum(c in vowels for c in b)
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->

solution/1700-1799/1704.Determine if String Halves Are Alike/README_EN.md

@@ -54,6 +54,14 @@ class Solution:
         return cnt == 0
 ```
 
+```python
+class Solution:
+    def halvesAreAlike(self, s: str) -> bool:
+        vowels = set('aeiouAEIOU')
+        a, b = s[:len(s) >> 1], s[len(s) >> 1:]
+        return sum(c in vowels for c in a) == sum(c in vowels for c in b)
+```
+
 ### **Java**
 
 ```java