# LeetCode: 235. Lowest Common Ancestor of a Binary Search Tree

## 题目¶

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).”

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.
```

Example 3:

```Input: root = [2,1], p = 2, q = 1
Output: 2
```

Constraints:

• The number of nodes in the tree is in the range [2, 10^5].
• -10^9 <= Node.val <= 10^9
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

## 解法¶

• None ，则返回 None
• 节点值等于 p 或 q 的值，则当前节点即为要找的 LCA，因为当前节点是 p 或 q 其中一个节点，不会有比它更近的共同祖先了。
• 节点值比 p 和 q 的值都大，根据 BST 的特性，改为从当前节点的左子数中查找
• 节点值比 p 和 q 的值都小，根据 BST 的特性，改为从当前节点的右子数中查找
• 节点值比其中一个大，比另一个小，根据 BST 的特性，p 和 q 分别分布在当前节点的左右子数中， 当前节点即为要找的 LCA。

```# 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):
if root is None:
return None

if root.val == p.val or root.val == q.val:
return root

if root.val > p.val and root.val > q.val:
return self.lowestCommonAncestor(root.left, p, q)
if root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right, p, q)

return root
```