# 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 ]))