235. Lowest Common Ancestor of a Binary Search Tree

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.

Constraints:

* `All of the nodes' values will be unique`.
* `p and q are different and both values will exist in the BST`.

### Analyze

1. if `p` node and `q` node are in `two` side of current iterator node `n`, the LCA is `n`;
2. if `p` node and `q` node are in `one` side of current iterator node `n`, to loop the one step in the other side node;

```js
/**
 * 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) {
  return lca(root, p, q)
};

var lca = function(node, p, q) {
  if (!node) return null
  if ((p.val - node.val) * (q.val - node.val) <= 0) {
    return node
  } else if (p.val - node.val < 0) {
    return lca(node.left, p, q)
  } else {
    return lca(node.right, p, q)
  }
}