Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes
q as the lowest node in
T that has both
q as descendants (where we allow a node to be a descendant of itself).”
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.
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.
Input: root = [2,1], p = 2, q = 1 Output: 2
2 important things to note:
- We start at the root, because an ancestor is always higher so must always start at the root.
- it's a binary search tree.
If we're looking to find
p = 2, it'll be in the left subtree because 2 is less than 6 (binary search tree property).
# 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 """ curr = root # p and q are gurnateed to exist in the tree # makes it run forever until we find the result while curr: # if p and q greater than root, it's right subtree if p.val > curr.val and q.val > curr.val: curr = curr.right elif p.val < curr.val and q.val < curr.val: # elif its left subtree curr = curr.left else: # we found the split where we have found our result return curr # no return stateemnt outside of loop as it's guranteed to exist # see the constraints # > p and q will exist in the BST.