【算法与数据结构】98、LeetCode验证二叉搜索树

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：注意不要落入下面你的陷阱，笔者本来想左节点键值<中间节点键值<右节点键值即可，写出如下代码：

class Solution2 {
public:// 1、输入参数， 返回值为result,以引用方式传入void traversal_preOrder(TreeNode* cur, bool &result) {// 2、终止条件if (cur == NULL) return;if (cur->left && cur->left->val >= cur->val) result = 0;if (cur->right && cur->right->val <= cur->val) result = 0;// 3、单层递归逻辑if (result) traversal_preOrder(cur->left, result);     // 左if (result) traversal_preOrder(cur->right, result);    // 右}bool isValidBST(TreeNode* root) {bool result = 1;traversal_preOrder(root, result);return result;}
};

  在leetcode执行时遇到下面的错误，再次读题，发现是所有左子树的键值小于中间节点键值，所有右子树键值大于中间节点键值，这个性质和中序遍历有所关联。如此一来，我们就想到用中序遍历来做，中序遍历数组是一个有序数组，判断该数组是否有序即可。

  程序如下：

class Solution {
public:void traversal_midOrder(TreeNode* cur, vector<int>& vec) {if (cur == NULL) return;traversal_midOrder(cur->left, vec);     // 左vec.push_back(cur->val);                // 中traversal_midOrder(cur->right, vec);    // 右}bool isValidBST(TreeNode* root) {if (root == NULL) return {};vector<int> v;traversal_midOrder(root, v);for (int i = 0; i < v.size()-1; i++) {if (v[i] >= v[i + 1]) return false;}return true;}
};

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:void traversal_midOrder(TreeNode* cur, vector<int>& vec) {if (cur == NULL) return;traversal_midOrder(cur->left, vec);     // 左vec.push_back(cur->val);                // 中traversal_midOrder(cur->right, vec);    // 右}bool isValidBST(TreeNode* root) {if (root == NULL) return {};vector<int> v;traversal_midOrder(root, v);for (int i = 0; i < v.size()-1; i++) {if (v[i] >= v[i + 1]) return false;}return true;}
};class Solution2 {
public:// 1、输入参数， 返回值为result,以引用方式传入void traversal_preOrder(TreeNode* cur, bool &result) {// 2、终止条件if (cur == NULL) return;if (cur->left && cur->left->val >= cur->val) result = 0;if (cur->right && cur->right->val <= cur->val) result = 0;// 3、单层递归逻辑if (result) traversal_preOrder(cur->left, result);     // 左if (result) traversal_preOrder(cur->right, result);    // 右}bool isValidBST(TreeNode* root) {bool result = 1;traversal_preOrder(root, result);return result;}
};// 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;
}int main()
{vector<string> t = { "5", "4", "NULL", "NULL", "6", "3", "NULL", "NULL", "7", "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:");Solution s;bool result = s.isValidBST(root);if (result) cout << "是一棵二叉搜索树" << endl;else cout << "不是一棵二叉搜索树" << endl;system("pause");return 0;
}

end