所有的LeetCode题解索引，可以看这篇文章——【算法和数据结构】LeetCode题解。

一、题目

二、解法

  思路分析：本题和这道题类似【算法与数据结构】236、LeetCode二叉树的最近公共祖先，相同的算法也能解这道题，但是没有充分利用到二叉搜索树的性质，二叉搜索树的性质为中间节点的键值大于所有左子树节点的键值，大于所有右子树的键值。因此要找到两个节点的最近祖先节点，只需要确定节点位于[p,q]（或者是[q,p]大小未知，左闭右闭的区间）区间内。那么当节点键值比两个节点的键值都小时，说明公共最先节点在右子树内，遍历右子树即可，反之亦然，最终剩下的就是最近公共祖先节点。
  程序如下：

class Solution {
public:// 后序遍历: 左右中// 1、输入参数TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {// 2、终止条件if (root == NULL) return root;    // 如果节点相等或者是空节点返回// 3、单层递归逻辑if ((root->val > q->val && root->val > p->val)) {TreeNode* left = lowestCommonAncestor(root->left, p, q);    // 左if (left != NULL) return left;}if ((root->val < q->val && root->val < p->val)) {TreeNode* right = lowestCommonAncestor(root->right, p, q);    // 右if (right != NULL) return right;}   return root;}
};

三、完整代码

# include <iostream>
# include <vector>
# include <string>
# include <queue>
using namespace std;// 树节点定义
struct TreeNode {int val;TreeNode* left;TreeNode* right;TreeNode() : val(0), left(nullptr), right(nullptr) {}TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}TreeNode(int x, TreeNode* left, TreeNode* right) : val(x), left(left), right(right) {}
};class Solution {
public:// 后序遍历: 左右中// 1、输入参数TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {// 2、终止条件if (root == NULL) return root;    // 如果节点相等或者是空节点返回// 3、单层递归逻辑if ((root->val > q->val && root->val > p->val)) {TreeNode* left = lowestCommonAncestor(root->left, p, q);    // 左if (left != NULL) return left;}if ((root->val < q->val && root->val < p->val)) {TreeNode* right = lowestCommonAncestor(root->right, p, q);    // 右if (right != NULL) return right;}   return root;}
};// 前序遍历迭代法创建二叉树，每次迭代将容器首元素弹出（弹出代码还可以再优化）
void Tree_Generator(vector<string>& t, TreeNode*& node) {if (!t.size() || t[0] == "NULL") return;    // 退出条件else {node = new TreeNode(stoi(t[0].c_str()));    // 中if (t.size()) {t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->left);              // 左}if (t.size()) {t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->right);             // 右}}
}template<typename T>
void my_print(T& v, const string msg)
{cout << msg << endl;for (class T::iterator it = v.begin(); it != v.end(); it++) {cout << *it << ' ';}cout << endl;
}template<class T1, class T2>
void my_print2(T1& v, const string str) {cout << str << endl;for (class T1::iterator vit = v.begin(); vit < v.end(); ++vit) {for (class T2::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {cout << *it << ' ';}cout << endl;}
}// 层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {queue<TreeNode*> que;if (root != NULL) que.push(root);vector<vector<int>> result;while (!que.empty()) {int size = que.size();  // size必须固定, que.size()是不断变化的vector<int> vec;for (int i = 0; i < size; ++i) {TreeNode* node = que.front();que.pop();vec.push_back(node->val);if (node->left) que.push(node->left);if (node->right) que.push(node->right);}result.push_back(vec);}return result;
}// 前序遍历,找二叉树中指定的键值
TreeNode* traversal_preOrder(TreeNode* cur, int val) {if (cur == NULL) return NULL;if (cur->val == val) return cur;        // 中if (traversal_preOrder(cur->left, val) != NULL) return traversal_preOrder(cur->left, val);        // 左   if (traversal_preOrder(cur->right, val) != NULL) return traversal_preOrder(cur->right, val);     // 右  return NULL;
}int main()
{// 构建二叉树vector<string> t = { "6", "2", "0", "NULL", "NULL", "4", "3", "NULL", "NULL", "5", "NULL", "NULL", "8", "7", "NULL", "NULL", "9", "NULL", "NULL" };   // 前序遍历my_print(t, "目标树");TreeNode* root = new TreeNode();Tree_Generator(t, root);vector<vector<int>> tree = levelOrder(root);my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");// 构建p, q节点int p = 2, q = 8;TreeNode* P_node = traversal_preOrder(root, p);TreeNode* Q_node = traversal_preOrder(root, q);// 找最近公共祖先节点Solution s;TreeNode* result = s.lowestCommonAncestor(root, P_node, Q_node);cout << "节点 " << P_node->val << " 和节点 " << Q_node->val << " 的最近公共祖先节点为 " << result->val << endl;system("pause");return 0;
}

end