NC102 在二叉树中找到两个节点的最近公共祖先（C++）- 中等、树

题目链接：https://www.nowcoder.com/practice/e0cc33a83afe4530bcec46eba3325116?tpId=196&rp=1&ru=%2Fexam%2Foj&qru=%2Fexam%2Foj&sourceUrl=%2Fexam%2Foj%3Fpage%3D2%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196&difficulty=&judgeStatus=&tags=&title=102&gioEnter=menu
题目如下：

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param o1 int整型 
     * @param o2 int整型 
     * @return int整型
     */
    int lowestCommonAncestor(TreeNode* root, int o1, int o2) {
        return dfs(root,o1,o2)->val;//在力扣236题的基础上，显示它的值
    }
    TreeNode* dfs(TreeNode* root,int val1,int val2){
        if(!root||root->val==val1||root->val==val2) return root;
        
        auto left=dfs(root->left,val1,val2);
        auto right=dfs(root->right,val1,val2);
        
        if(left&&right) return root;//返回的当前节点的父节点
        
        return left!=nullptr?left : right;//逻辑无意义，只语法意义
    }
};
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