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.

/**
 * 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) {
        TreeNode *result = NULL;
	    findLCA(root, p, q, &result);

	    return result;
    } 
private:
    bool findNode(TreeNode *root, TreeNode *p)
    {
	    if (root == NULL || p == NULL)
	    {
		    return false;
	    }

	    if (root == p)
	    {
		    return true;
	    }

	    return findNode(root->left, p) || findNode(root->right, p);
    }

    int findLCA(TreeNode *root, TreeNode *p, TreeNode *q, TreeNode **result)
    {
	    if (root == NULL)
	    {
		    return 0;
	    }

	    if (root == p)
	    {
		    if (findNode(root, q))
		    {
			    *result = root;
			    return 2;
		    }
		    else
		    {
			    return 1;
		    }
	    }
	    else if (root == q)
	    {
		    if (findNode(root, p))
		    {
			    *result = root;
			    return 2;
		    }
		    else
		    {
			    return 1;
		    }
	    }
	    else 
	    {
		    int left = findLCA(root->left, p, q, result);
		    int right = 0;
		    if (left != 2)
		    {
			    right = findLCA(root->right, p, q, result);
		    }
		    if (left == 1 && right == 1)
		    {
			    *result = root;
		    }
		    return left + right;
	    }
    }
};