Recent common ancestor binary search tree

Leecode brush title

• Title Description

Given a binary search tree, find the nearest common ancestor of the two specified nodes of the tree.
Baidu Encyclopedia recent common ancestor as defined: "For two nodes of the root of the tree T p, q, is represented as a common ancestor node x, that x is p, q ancestor depth as large as possible and x (a node can be its own ancestors). "

• Example
  For example, given the following binary search tree: root = [6,2,8,0,4,7,9, null, null, 3,5]
  

Input: root = [6,2,8,0,4,7,9, null, null, 3,5], p = 2, q = 8
Output: 6
Explanation: nodes 2 and 8 is the common ancestor 6.
Input: root = [6,2,8,0,4,7,9, null, null, 3,5], p = 2, q = 4
Output: 2
Explanation: nodes 2 and 4 are common ancestor 2, because by definition a recent common ancestor node can be a node itself.

• Code

//注意利用搜索二叉树的特性即可
/**
 * 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(p->val>=root->val&&q->val<=root->val||p->val<=root->val&&q->val>=root->val)
        {
            return root;
        }
        else if(p->val>=root->val)
        {
            TreeNode* node;
            node = lowestCommonAncestor(root->right,p,q);
            return node;
        }
        else if(p->val<=root->val)
        {
            TreeNode* node;
            node = lowestCommonAncestor(root->left,p,q);
            return node;
        }
        else
            return NULL;
    }
};