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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              
    ___2__          ___8__
   /              /      
   0      _4       7       9
         /  
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Analyse: For a binary search tree, we can recursively compare the node value and p, q value. 

Trial 1: comparison

Runtime: 44ms.

/**
 * 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) return NULL;
        if(root == p || root == q) return root;
        if(root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right, p, q);
        if(root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left, p, q);
        return root;
    }
};

Trial 2: same solution as Lowest Common Ancestor. 

Runtime: 44ms.

/**
 * 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) return NULL;
        if(root == p || root == q) return root;
        TreeNode* leftSub = lowestCommonAncestor(root->left, p, q);
        TreeNode* rightSub = lowestCommonAncestor(root->right, p, q);
        if(leftSub && rightSub) return root;
        return leftSub ? leftSub : rightSub;
    }
};