面试题68 - I：二叉搜索树的最近公共祖先（C++）

/**
 * 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* cur = root;
        while(cur != nullptr)
        {
            if(cur->val > p->val && cur->val > q->val)
            {
                cur = cur->left; //节点p和节点q均在节点cur的左子树中，继续向左遍历
            }
            else if(cur->val val && cur->val val)
            {
                cur = cur->right; //节点p和节点q均在节点cur的右子树中，继续向右遍历
            }
            else
                break; //节点p和节点q均在节点cur的两侧，跳出循环，找到了最近公共祖先节点cur
        }
        return cur;
    }
};

/**
 * 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 == nullptr) return nullptr;
        if(root->val > p->val && root->val > q->val)
        {
            return lowestCommonAncestor(root->left, p, q);
        }
        if(root->val val && root->val val)
        {
            return lowestCommonAncestor(root->right, p, q);
        }
        return root;
    }
};