Lowest Common Ancestor of a Binary Search Tree
Created: November 1, 2018 by [lek-tin]
Last updated: November 1, 2018
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.
Solution
Recursion
Time complexity: O(n)
Space complexity: O(n)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
parentVal = root.val
pVal, qVal = p.val, q.val
if pVal > parentVal and qVal > parentVal:
return self.lowestCommonAncestor(root.right, p, q)
elif pVal < parentVal and qVal < parentVal:
return self.lowestCommonAncestor(root.left, p, q)
else:
return root
DFS using stack
Time complexity: O(n)
Space complexity: O(n)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
pVal, qVal = p.val, q.val
stack = [root]
while stack:
node = stack.pop()
if node:
parentVal = node.val
if pVal > parentVal and qVal > parentVal:
stack.append(node.right)
elif pVal < parentVal and qVal < parentVal:
stack.append(node.left)
else:
return node