# How to solve Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

## Finding the lowest common ancestor node of two given nodes in a binary search tree

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### Problem statement

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.

### Solution: Recursion

Note that in a BST, the values of a node and its children left and right satisfy

left.value < node.value < right.value.

It lets you know which branch (left or right) of the root the nodes p and q belong to.

#### Code

#include <iostream>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
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 (root->val < p->val && root->val < q->val) {
return lowestCommonAncestor(root->right, p, q);
}
return root;
}

int main() {
TreeNode zero(0);
TreeNode three(3);
TreeNode five(5);
TreeNode four(4);
four.left = &three;
four.right = &five;
TreeNode two(2);
two.left = &zero;
two.right = &four;
TreeNode seven(7);
TreeNode nine(9);
TreeNode eight(8);
eight.left = &seven;
eight.right = &nine;
TreeNode six(6);
six.left = &two;
six.right = &eight;

cout << lowestCommonAncestor(&six, &two, &eight)->val << endl;
cout << lowestCommonAncestor(&six, &two, &four)->val << endl;
cout << lowestCommonAncestor(&two, &two, &zero)->val << endl;
}
Output:
6
2
2

#### Complexity

• Runtime: O(logN) (the height of the tree), where N is the number of nodes.

• Extra space: O(1).