题解 | NO.37#二叉搜索树的最近公共祖先#3.13
二叉搜索树的最近公共祖先
https://www.nowcoder.com/practice/d9820119321945f588ed6a26f0a6991f
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ /** * 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(struct TreeNode* root, int p, int q ) { struct TreeNode* cur = root; int temp = 0; //将p指向较小的值,q指向较大的值 if (q < p ) { temp = p; p = q; q = temp; } while (1) { //p为最近公共祖先 if (cur->val == p) return p; //q为最近公共祖先 else if (cur->val == q) return q; //cur所指结点为最近公共祖先 else if (cur->val > p && cur->val < q) return cur->val; //最近公共祖先在cur的左子树 else if (cur->val > p && cur->val > q) cur = cur->left; //最近公共祖先在cur的右子树 else cur = cur->right; } } //最近公共祖先一定是在p、q之间的值