235. Lowest Common Ancestor of a Binary Search Tree

1. Description

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).”

2. Example

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

3. Note

• 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.

4. Solutions

My Accepted Solution

n is number of nodes in the i_root
Time complexity: O(lgn)
Space complexity: O(1)

// it is easy to turn the iteration method into a recursive methord, but it is meaningless

class Solution 
{
public:
    // TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    TreeNode* lowestCommonAncestor(TreeNode *i_root, TreeNode *i_nodeP, TreeNode *i_nodeQ) 
    {
        auto result = i_root;
        while(true)
        {
            if(i_nodeP->val < result->val && i_nodeQ->val < result->val)
                result = result->left;
            else if(i_nodeP->val > result->val && i_nodeQ->val > result->val)
                result = result->right;
            else
                break;
        }
        
        return result;
    }
};