力扣-98题验证二叉搜索树（C++）- 我家大宝的三个解法+我的一个解法

秒速五厘米 2022-09-01 13:47 120阅读 0赞

题目链接：[https://leetcode-cn.com/problems/validate-binary-search-tree/][https_leetcode-cn.com_problems_validate-binary-search-tree]  
题目如下：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70]  
注：一二为递归，三四为非递归

解法一：（大宝的）

/**
     * Definition for a binary tree node.
     * 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:
        long long pre=LONG_MIN;
        bool flag=true;
        
        void dfs(TreeNode* root) {
        
            if (!root) return;
            if (!flag) return;
    
            dfs(root->left);
    
            if (root->val <= pre) flag=false;
            pre=root->val;
    
            dfs(root->right);
    
        }
        
        bool isValidBST(TreeNode* root) {
        
            //conclusion:二叉搜索树的中序遍历都是有序的
            
            dfs(root);
            return flag;
    
        }
    };

解法二：（大宝的）

/**
     * Definition for a binary tree node.
     * 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:
        long long pre=LONG_MIN;
        
        bool dfs(TreeNode*root) {
        
            if (!root) return true;
    
    		//如果左子树为空，则不执行该语句
            if (!dfs(root->left)) {
         //如果返回true，则为false，不执行语句
                return false;
            }
    
            if (root->val <= pre) return false;//二叉搜索树中，pre即左孩子应<root->val
            else pre=root->val;
    
            return dfs(root->right);
        }
    
        bool isValidBST(TreeNode* root) {
        
            //conclusion:二叉搜索树的中序遍历都是有序的
    
            return dfs(root);
    
        }
    };

解法三：（我的）

/**
     * Definition for a binary tree node.
     * 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:
        bool isValidBST(TreeNode* root) {
        
            //结论：二叉搜索树的中序遍历都是有序的
            vector<int> inorder;
            stack<TreeNode*> stk;
    
            while(root||stk.size()){
        
                while(root){
        
                    stk.push(root);
                    root=root->left;
                }
    
                root=stk.top();
                stk.pop();
                inorder.push_back(root->val);
                root=root->right;
            }
    
            for(int i=0;i<inorder.size()-1;i++){
        
                if(inorder[i]>=inorder[i+1]) return false;
            }
    
            return true;
    
        }
    };

解法四：（大宝的）

/**
     * Definition for a binary tree node.
     * 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:
        bool isValidBST(TreeNode* root) {
        
            //conclusion:二叉搜索树的中序遍历都是有序的
    
            vector<int> inorder;
            stack<TreeNode*> stk;
            long pre=LONG_MIN;
            while(root||stk.size()){
        
                while(root){
        
                    stk.push(root);
                    root=root->left;
                }
    
                auto t=stk.top();
                stk.pop();
                if (t->val <= pre) return false;
                pre=t->val;
    
                root=t;
                root=root->right;
            }
            return true;
        }
    };

[https_leetcode-cn.com_problems_validate-binary-search-tree]: https://leetcode-cn.com/problems/validate-binary-search-tree/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70]: /images/20220829/95af154f3a83420bab598a785be15489.png