235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) node 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, 10^5].
• -10^9 <= Node.val <= 10^9
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

Idea

递归思路：
根据二叉查找树的定义：左子树的子树的所有节点值都小于根节点，右子树的所有节点值都大于根节点
* 如果p,q节点值都比当前节点都小，则说明p,q 都在当前节点的左子树中
* 如果p,q节点值都比当前节点都大，则说明p,q都在当前节点的右子树中
* 如果p,q当中任意一个节点值等于当前节点，则说明当前节点即为p,q公共最近祖先

JavaScript 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) {
    
    if( root.val > p.val && root.val > q.val ){
        return lowestCommonAncestor(root.left, p, q)
    }
    if( root.val < p.val && root.val < q.val ){
        return lowestCommonAncestor(root.right, p, q)
    }
    return root;
};