【leetcode】Validate Binary Search Tree

Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

思路一：利用递归的思想，任何一个节点必须要在min和max范围内

 

min和max根据是左子树还是右子树来确定

 

如果是左子树：

node->left<node

 

如果是右子树：

node->right>node

 

 

递归条件：node->val在min和max范围内，node->left要在min,node->val范围内，node->right要在node->val,max范围内

node->val<max&&node->val>min&&testValid(node->left,node->val,min)&&testValid(node->right,max,node->val); 

 

 

 

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     bool isValidBST(TreeNode *root) {

         if(root==NULL)
         {
             return true;
         }

         if(root->left==NULL&&root->right==NULL)
         {
             return true;
         }

         //注意如果测试用例含有INT_MAX则，必须采用long long int 才能避免出错
         return testValid(root,(long long int)INT_MAX+,(long long int)INT_MIN-);
     }

     bool testValid(TreeNode *node,long long int max, long long int min)
     {
         if(node==NULL)
         {
             return true;
         }

         return node->val<max&&node->val>min&&testValid(node->left,node->val,min)&&testValid(node->right,max,node->val);

     }
 };

 

 

 

第二种方法，利用 二叉排序树 中序遍历 结果为递增序列的性质

 

 

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     vector<int> ret_v;
     bool isValidBST(TreeNode *root) {

         if(root==NULL)
         {
             return true;
         }

         ret_v.clear();
         inOrderTraversal(root);

         //判断是否递增
         for(int i=;i<ret_v.size()-;i++)
         {
             if(ret_v[i]>=ret_v[i+])
             {
                 return false;
             }
         }
         return true;
     }

     //中序遍历，记录下数值
     void inOrderTraversal(TreeNode *root)
     {
         if(root==NULL)
         {
             return;
         }

         inOrderTraversal(root->left);
         ret_v.push_back(root->val);
         inOrderTraversal(root->right);
     }
 };

 

 

 

第三种：直接在中序遍历中判断：

 

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:

     int ret;
     bool flag;
     bool isFirstNode;

     bool isValidBST(TreeNode *root) {

         flag=true;
         //判断是不是第一个被访问的节点
         isFirstNode=true;

         inOrderTraversal(root);
         return flag;
     }

     void inOrderTraversal(TreeNode *root)
     {
         if(root==NULL)
         {
             return;
         }

         if(!flag)
         {
             return;
         }

         inOrderTraversal(root->left);

         //一旦发现不符合升序，则不是二叉排序树
         if(!isFirstNode&&ret>=root->val)
         {
             flag=false;
             return;
         }

         ret=root->val;
         isFirstNode=false;

         inOrderTraversal(root->right);
     }

 };