Lowest Common Ancestor of a Binary Search Tree

Problem

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Solution

Recursive Solution, Easy to come up with

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

Iterative Solution, Use Difference of Val

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == null || q == null) return null;
        while ((root.val - p.val) * (root.val - q.val) > 0) {
            root = root.val < p.val ? root.right : root.left;
        }
        return root;
    }
}

Analysis

Don't get panic because of the long question's name
A pretty straightforward problem if thinking in recursion
In both solutions above, we continue to visit the nodes if p and q are in the same side of root
Otherwise, we simply return the current root cause there is no common nodes anymore in the below

Complexity:

• Time: O(n), since we have to visit all nodes in the worst case
• Space: O(1) for iterative solution, O(n) for recursive solution