lowest common ancestor of a binary search tree easy

Tree/LowestCommonAncestorOfABinarySearchTree.py

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+"""
+Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
+
+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 binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]
+
+        _______6______
+       /              \
+    ___2__          ___8__
+   /      \        /      \
+   0      _4       7       9
+         /  \
+         3   5
+Example 1:
+
+Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
+Output: 6
+Explanation: The LCA of nodes 2 and 8 is 6.
+Example 2:
+
+Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
+Output: 2
+Explanation: The LCA of nodes 2 and 4 is 2, 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 BST.
+
+
+给定一颗 BST，找到两个子节点的最小公共祖先。
+用普通的树的方法也是可以的，不过可以做个剪枝优化。
+
+因为 BST 我们知道每个节点的左右子节点的范围。
+
+1. 如果处于 root 左右，那么直接返回即可。
+2. 如果都小于，那么去找左子树。
+3. 如果都大，那么去找右子树。
+
+在寻找过程中，
+4. 只要有一个命中了，那么直接返回当前节点即可。因为剩下的那个节点只有可能在它的子树中，如果不在它的子树中也不会执行到这一步。
+
+beat 99%
+
+测试地址：
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/
+
+"""
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def lowestCommonAncestor(self, root, p, q):
+        """
+        :type root: TreeNode
+        :type p: TreeNode
+        :type q: TreeNode
+        :rtype: TreeNode
+        """
+
+        if p.val == root.val or q.val == root.val:
+            return root
+
+        if p.val < root.val and q.val > root.val:
+            return root
+        elif p.val > root.val and q.val < root.val:
+            return root
+
+        if p.val > root.val and q.val > root.val:
+
+            return self.lowestCommonAncestor(root.right, p, q)
+
+        else:
+            return self.lowestCommonAncestor(root.left, p, q)
+