题解 | #二叉搜索树的最近公共祖先#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @param p int整型 
     * @param q int整型 
     * @return int整型
     */
    int lowestCommonAncestor(TreeNode* root, int p, int q) {
        // write code here
        if(root == nullptr){
            return -1;
        }
        if(root->val == p || root->val == q){
            return root->val;
        }
        
        int left = lowestCommonAncestor(root->left, p, q);
        int right = lowestCommonAncestor(root->right, p, q);
        
        if(left > -1 && right > -1){
            return root->val;
        }
        if(left == -1 && right == -1){
            return -1;
        }
        
        return left == -1 ? right : left;
    }
    
    
};