Lowest Common Ancestor of a Binary Search Tree.md

Notes from Problem Statement

• "A node is a descendant of itself"
• "All node values are unique"
• "p and q are different and both values will exist in the BST"

Algorithm

• Notice we have a binary search tree. That means the numbers are ordered.
• Traverse down the tree. At each node:
  • if the node's value is smaller than both p and q, go right
  • if it's bigger, go left
  • otherwise, we've found the Lowest Common Ancestor

Provided Code

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
}

Solution

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {   
        if (root == null || p == null || q == null) {
            return null;
        }
        TreeNode node = root;
        while (node != null) {
            if (node.val < p.val && node.val < q.val) {
                node = node.right;
            } else if (node.val > p.val && node.val > q.val) {
                node = node.left;
            } else {
                return node;
            }
        }
        return null;
    }
}

Time/Space Complexity

• Time Complexity: O(log n) if balanced tree, O(n) otherwise
• Space Complexity: O(1)

Links

• github.com/RodneyShag