# Leetcode: Lowest Common Ancestor of a Binary Search Tree

Lowest Common Ancestor of a Binary Search Tree

Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/lowest-common-ancestor-of-a-binary-search-tree
## Basic Ideas: For BST, get min(p.val, q.val) and max(p.val, q.val)
##              Check from the root node
##              If both are smaller than root.val, move the left sub-tree
##              If both are bigger than root.val, move the right-tree
##              If one smaller and one bigger, the current node is what we want
## Assumption: No duplicate value in the BST
## Complexity: Time O(k), Space O(1). k is the height
# 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
"""
stack = []
r = root
min_val = min(p.val, q.val)
max_val = max(p.val, q.val)
while r and (r.val > max_val or r.val < min_val):
if r.val > max_val:
r = r.left
else:
r = r.right
return r
```

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