235. Lowest Common Ancestor of a Binary Search Tree.js

/**
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).”

Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5
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.
Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the BST.
 */
/**
 * Leetcode fundamental: 11/5 Update
 * Failure:
 * 1. Fail to think of condition to find next child node by comparing p.val, q.val and curr.val
 * 
 * Note: Hard to think of stack and while loop solution
 * Memoize the solution!
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
  // Condition to find left child node (p.val < currNode.val && q.val < currNode.val)
  if (p.val < root.val && q.val < root.val)
    return lowestCommonAncestor(root.left, p, q);
  // Codition to find right child node (p.val > currNode.val && q.val > currNode.val)
  if (p.val > root.val && q.val > root.val)
    return lowestCommonAncestor(root.right, p, q);
  // Other case
  else return root;
};