# 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: β€œThe 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;
    }
};