 Time Flying  ### 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 = 8Output: 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 = 4Output: 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)  }}