0235. Lowest Common Ancestor of a Binary Search Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree

Description

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

**Input:** root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
**Output:** 6
**Explanation:** The LCA of nodes 2 and 8 is 6.

Example 2:

**Input:** root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
**Output:** 2
**Explanation:** The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

**Input:** root = [2,1], p = 2, q = 1
**Output:** 2

Constraints:

• The number of nodes in the tree is in the range [2, 105].

• -109 <= Node.val <= 109

• All Node.val are unique.

• p != q

• p and q will exist in the BST.

ac

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // exit
        if (root == null || root.val == p.val || root.val == q.val) return root;

        // p/q in different subtree
        if (p.val < root.val && root.val < q.val || q.val < root.val && root.val < p.val) {
            return root;
        } 
        // p/q in same subtree, recursion
        else if (p.val < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        } else {
            return lowestCommonAncestor(root.right, p, q);
        }
    }
}

/*
1)p/q in different subtree, return root; 2) p/q in same subtree, recursion. if root.val == p.val || root.val == q.val return current root, root must be the answer
*/