题解 | #判断是不是二叉搜索树#

1.中序遍历递归

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 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 
     * @return bool布尔型
     */
    void inorder(TreeNode* root,vector<int> &res)
    {
        if(!root)
            return;
        inorder(root->left, res);
        res.push_back(root->val);
        inorder(root->right, res);
    }
    bool isValidBST(TreeNode* root) {
        if(!root)
            return true;
        if(!isValidBST(root->left))
            return false;
        if(root->val <= pre)
            return false;
        pre = root->val;
        if(!isValidBST(root->right))
            return false;
        return true;
    }
};

2.获取中序遍历的结果然后检查是否严格递增

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 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 
     * @return bool布尔型
     */
    void inorder(TreeNode* root,vector<int> &res)
    {
        if(!root)
            return;
        inorder(root->left, res);
        res.push_back(root->val);
        inorder(root->right, res);
    }
    long pre = INT_MIN;
    bool isValidBST(TreeNode* root) {
        // write code here
        vector<int> res;
        inorder(root, res);
        if(res.size() <= 1)
            return true;
        for(int i = 1;i<res.size();++i)
        {
            if(res[i] <= res[i-1])
                return false;
        }
        return true;
    }
};

class Solution {
public:
	bool isValidBST(TreeNode* root) {
	    //使用栈获取中序遍历的结果
		stack<TreeNode*> s;
        TreeNode* head = root;
        vector<int> res; //记录中序遍历的数组
        while(head != NULL || !s.empty()){
            //直到没有左节点
            while(head != NULL){  
                s.push(head);
                head = head->left;
            }
            head = s.top();
            s.pop();
            //访问节点
            res.push_back(head->val);
            head = head->right;
        }
        if(res.size() <= 1)
            return true;
        for(int i = 1;i<res.size();++i)
        {
            if(res[i] <= res[i-1])
                return false;
        }
        return true;
	}
};