Don't overthink: just cache it
I still feel that a lot of times programmers overthink solutions too much.
They think about a complex algorithm to save few bytes when a simpler one with a little bit of extra memory would do the job just fine. They think about building ML models when a simple rule-based approach would do the job just fine. They think about building solutions that will solve the world's hunger problem - honorable, but honestly it would be better for everyone if they just focused on solving the smaller problem first.
A rule of thumb that works in many more cases than you might imagine is this:
Check this problem out: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree-iii/
1650. Lowest Common Ancestor of a Binary Tree III
Medium
Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).
Each node will have a reference to its parent node. The definition for Node is below:
class Node {
public int val;
public Node left;
public Node right;
public Node parent;
}
According to the definition of LCA on Wikipedia : "The lowest common ancestor of two nodes p and q in a tree T is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself)."
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1 Output: 3 Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4 Output: 5 Explanation: The LCA of nodes 5 and 4 is 5 since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [1,2], p = 1, q = 2 Output: 1
Constraints:
- The number of nodes in the tree is in the range
[2, 105]. -109 <= Node.val <= 109- All
Node.valare unique. p != qpandqexist in the tree.
Accepted
284
Submissions
365
You can literally solve it in 5 lines of code: traverse one path up and cache it. Traverse the other, looking for a match in the cache. That's it.
Why even bother thinking about more complicated solutions than that? As I said many times, memory is cheap - time isn't. Code is below, cheers, ACC.
public Node LowestCommonAncestor(Node p, Node q)
{
if (p == null || q == null) return null;
Hashtable path = new Hashtable();
for (Node temp = p; temp != null; temp = temp.parent) path.Add(temp.val, true);
for (Node temp = q; temp != null; temp = temp.parent) if (path.ContainsKey(temp.val)) return temp;
return null;
}
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