235. 二叉搜索树的最近公共祖先(leetcode)

https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/

明确函数的返回值和参数,返回最近公共祖先,没有就返回null

单层递归逻辑是:若按搜索树条件,递归左子树未查询到,递归右子树也未查询到,那么意味着root即是答案,直接返回,这样就完成了递归返回值的编写

/**
 * 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:
    
    // 返回公共祖先节点,若没有则返回null
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==nullptr)return root;
        if(root->val > p->val && root->val > q->val)
        {
            TreeNode* left = lowestCommonAncestor(root->left,p,q);
            if(left!=nullptr)return left;
        }
        if(root->val < p->val && root->val < q->val)
        {
            TreeNode* right = lowestCommonAncestor(root->right,p,q);
            if(right!=nullptr)return right;
        }
        // root在p和q中间
        return root;
    }
};