# Question

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 v and w as the lowest node in T that has both v and w as descendants (where we allow **a node to be a descendant of itself**).”

```
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
```

For example, the lowest common ancestor (LCA) of nodes `2`

and `8`

is `6`

. Another example is LCA of nodes `2`

and `4`

is `2`

, since a node can be a descendant of itself according to the LCA definition.

# Solution

```
# 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
"""
p_path = self.getPath(root, p)
q_path = self.getPath(root, q)
lca = root
for i in range(1, min(len(p_path), len(q_path))):
if p_path[i].val == q_path[i].val:
lca = p_path[i]
else:
break
return lca
def getPath(self, root, p):
if p.val == root.val:
return [ root ]
node_stack = []
node_stack.append((root, [ root ]))
while len(node_stack) != 0:
(node, path) = node_stack[-1]
if node.val == p.val:
return path
node_stack = node_stack[:-1]
if node.left != None:
node_stack.append((node.left, path + [ node.left ]))
if node.right != None:
node_stack.append((node.right, path + [ node.right ]))
```