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).”

&nbsp;
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


&nbsp;
Constraints:


	The number of nodes in the tree is in the range [2, 105].
	-109 &lt;= Node.val &lt;= 109
	All Node.val are unique.
	p != q
	p and q will exist in the BST.


class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
          
        if ((root.val >= p.val) and (root.val <= q.val)) or ((root.val >= q.val) and (root.val <= p.val)):
            return root
        elif (root.val > p.val):
            return self.lowestCommonAncestor(root.left, p, q)
        else:
            return self.lowestCommonAncestor(root.right, p , q)
/**
 * 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) {
        if(root == null || root == p || root == q)return root;

        TreeNode left = lowestCommonAncestor(root.left, p , q);
        TreeNode right = lowestCommonAncestor(root.right, p ,q);

        if(left == null)return right;
        if(right == null)return left;
        else{
            return root;
        }
    }
}
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p->val<root->val && q->val<root->val) return lowestCommonAncestor(root->left,p,q);
        else if(p->val>root->val && q->val>root->val) return lowestCommonAncestor(root->right,p,q);
        return root;
    }
};
var lowestCommonAncestor = function(root, p, q) {
    while(root){
        if(p.val>root.val && q.val>root.val){
            root=root.right
        }else if(p.val<root.val && q.val<root.val){
            root=root.left
        }else{
            return root
        }
    }
}
Lowest Common Ancestor of a Binary Search Tree