ALGODAILY

May 30, 2023

“It's lonely at the top. Ninety-nine percent of people in the world are convinced they are incapable of achieving great things, so they aim for the mediocre. The level of competition is thus fiercest for 'realistic' goals, paradoxically making them the most time and energy-consuming.”

- Tim Ferriss

Day 9: Lowest Common Ancestor

In the official React documentation, it's recommended that to find out where state data should live, the following steps be taken:

• Identify every component that renders something based on that state.
• Find a common owner component (a single component above all the components that need the state in the hierarchy).
• Either the common owner or another component higher up in the hierarchy should own the state.

This is an algorithmic problem-- let's find the lowest common ancestor!

You're given a binary search tree tree1 and two of its child nodes as parameters. Can you write a method lowestCommonAncestor(root: Node, node1: Node, node2: Node) to identify the lowest common ancestor of the two nodes? The lowest common ancestor is the deepest node that has both of the two nodes as descendants.

In the below example, the lowest common ancestor of node 5 and 8 is 7.

/*
       7
      / \
     4   8
    / \
   1   5
*/

The method will called in the following manner: lowestCommonAncestor(root, node1, node2);.

You can assume our standard node definition for the tree vertices:

// Node definition
function Node(val) {
  this.val = val;
  this.left = this.right = null;
}

# Node definition
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

Bonus: is there a way to find the lowest common ancestor of two nodes if the root wasn't passed as a parameter?

Constraints

• Number of nodes in the BST <= 1000
• The nodes will always contain integer values between -1000000000 and 1000000000
• Expected time complexity : O(n)
• Expected space complexity : O(1)

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