LeetCode 235. Lowest Common Ancestor of a Binary Search Tree (Algorithm Explained)

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Hey! There is another problem on leetcode which is literally the same exact thing but not a 'binary SEARCH tree', its considered Medium difficulty and I came up with a really inefficient solution haha. Would love to see you explain that since the tree nodes are not sorted. I find your videos really helpful bro, especially this time of year. Keep it up, looking forward to great content!

jaspreets

wow! that's easy! Thank you
Python:

class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':

if p.val < root.val and q.val < root.val:
return self.lowestCommonAncestor(root.left, p, q)

if p.val > root.val and q.val > root.val:
return self.lowestCommonAncestor(root.right, p, q)

return root

hassansmallah

That's awesome! Watching your videos has become my new daily routine.

ziqixu

bro everytime you explain a solution it sounds so easy 😭GOAT

pealhasan

runtime error: member access within null pointer of type 'TreeNode'

dunkyzhang

i think there is a problem, if p and q is 0, and 5 in the first example then it would just return the root?which is not correct

levente

Your explanation and problem solving is great. I would also like to know the time complexity and space complexity with explanation as these will be expected by any interviewer. :)

nagarjunareddypadala

You treat problems as if they are nothing. I like that attitude so much. Makes any problem easier.

rishabhjain

where is base case if root is null and why its not there?

AbhishekSharma-kcic

What will happen in the case of two nodes directly connected to each other?

anmolbindroo

This is so beautiful Nick..Thanks for the explanation.

somyasrivastava

some test cases are failing on leetcode for this solution

harenderbhardwaj

This will throw null pointer exception if root itself is null

sharmadenesh

this solution will create null pointer exception.

samspeaks-hkvp

Correction in the code :
if(p->val == root->val || q->val == root->val)
return root;

This will be the first thing to check. Thanks for the solution @Nick White

AniketSomwanshi-llmz

This solution only works for the Binary Search Tree (BST) and not for the General Binary Tree.

prashantsah

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