feat: azl397985856#236 add python 🐍 solution (azl397985856#278)

Author: jojoxiaojing

problems/236.lowest-common-ancestor-of-a-binary-tree.md

@@ -70,60 +70,11 @@ p and q are different and both values will exist in the binary tree.
 
 ## 代码
 
-```js
+代码支持： JavaScript， Python3
 
+- JavaScript Code:
 
-/*
- * @lc app=leetcode id=236 lang=javascript
- *
- * [236] Lowest Common Ancestor of a Binary Tree
- *
- * https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
- *
- * algorithms
- * Medium (35.63%)
- * Total Accepted:    267.3K
- * Total Submissions: 729.2K
- * Testcase Example:  '[3,5,1,6,2,0,8,null,null,7,4]\n5\n1'
- *
- * Given a binary tree, find the lowest common ancestor (LCA) of two given
- * nodes in the tree.
- *
- * According to the definition of LCA on Wikipedia: “The lowest common ancestor
- * is defined between two nodes p and q as the lowest node in T that has both p
- * and q as descendants (where we allow a node to be a descendant of itself).”
- *
- * Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]
- *
- *
- *
- * Example 1:
- *
- *
- * Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
- * Output: 3
- * Explanation: The LCA of nodes 5 and 1 is 3.
- *
- *
- * Example 2:
- *
- *
- * Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
- * Output: 5
- * Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant
- * of itself according to the LCA definition.
- *
- *
- *
- *
- * Note:
- *
- *
- * All of the nodes' values will be unique.
- * p and q are different and both values will exist in the binary tree.
- *
- *
- */
+```js
 /**
  * Definition for a binary tree node.
  * function TreeNode(val) {
@@ -147,6 +98,32 @@ var lowestCommonAncestor = function(root, p, q) {
 };
 ```
 
+- Python Code:
+
+``` 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 lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        if not root or root == p or root == q:
+            return root
+        left = self.lowestCommonAncestor(root.left, p, q)
+        right = self.lowestCommonAncestor(root.right, p, q)
+        
+        if not left:
+            return right
+        if not right:
+            return left
+        else:
+            return root
+
+```
+
 ## 扩展
 如果递归的结束条件改为`if (!root || root.left === p || root.right === q) return root;` 代表的是什么意思，对结果有什么样的影响？