## # 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]

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.

### # Solution

Approach 1: Recursion.

Approach 2: Iteration (no stack needed).

### # Code (Python)

Approach 1:

``````    def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if (root.val - p.val) * (root.val - q.val) <= 0:
return root
if root.val > p.val:
return self.lowestCommonAncestor(root.left, p, q)
else:
return self.lowestCommonAncestor(root.right, p, q)
``````

Approach 2:

``````    def lowestCommonAncestor(self, root, p, q):
while True:
if (root.val - p.val) * (root.val - q.val) <= 0:
return root
if root.val > p.val:
root = root.left
else:
root = root.right
``````

### # Code (C++)

Approach 1:

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == NULL) return NULL;
if (root->val > p->val && root->val > q->val)
return lowestCommonAncestor(root->left, p, q);
else if (root->val < p->val && root->val < q->val)
return lowestCommonAncestor(root->right, p, q);
return root;
}
};

// For general binary tree.
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root || root->val == p->val || root->val == q->val)
return root;
TreeNode *left = lowestCommonAncestor(root->left, p, q);
TreeNode *right = lowestCommonAncestor(root->right, p, q);
if (left && right)
return root;
else if (!left)
return right;
else
return left;
}
};
``````

Approach 2:

``````class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode *node = root;
while (node != NULL)
{
if (node->val > p->val && node->val > q->val)
node = node->left;
else if (node->val < p->val && node->val < q->val)
node = node->right;
else
return node;
}
return NULL;
}
};
``````