【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); 

 

 

 

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     bool isValidBST(TreeNode *root) {

13        

14         if(root==NULL)

15         {

16             return true;

17         }

18        

19         if(root->left==NULL&&root->right==NULL)

20         {

21             return true;

22         }

23        

24         //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

25         return testValid(root,(long long int)INT_MAX+1,(long long int)INT_MIN-1);

26     }

27    

28    

29     bool testValid(TreeNode *node,long long int max, long long int min)

30     {

31         if(node==NULL)

32         {

33             return true;

34         }

35        

36        

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

38        

39     }

40 };

 

 

 

 

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

 

 

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12     vector<int> ret_v;

13     bool isValidBST(TreeNode *root) {

14        

15         if(root==NULL)

16         {

17             return true;

18         }

19        

20         ret_v.clear();

21         inOrderTraversal(root);

22        

23         //判断是否递增

24         for(int i=0;i<ret_v.size()-1;i++)

25         {

26             if(ret_v[i]>=ret_v[i+1])

27             {

28                 return false;

29             }

30         }

31         return true;

32     }

33    

34     //中序遍历，记录下数值

35     void inOrderTraversal(TreeNode *root)

36     {

37         if(root==NULL)

38         {

39             return;

40         }

41        

42         inOrderTraversal(root->left);

43         ret_v.push_back(root->val);

44         inOrderTraversal(root->right);

45     }

46 };

 

 

 

 

第三种：直接在中序遍历中判断：

 

 1 /**

 2  * Definition for binary tree

 3  * struct TreeNode {

 4  *     int val;

 5  *     TreeNode *left;

 6  *     TreeNode *right;

 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 8  * };

 9  */

10 class Solution {

11 public:

12    

13     int ret;

14     bool flag;

15     bool isFirstNode;

16  

17     bool isValidBST(TreeNode *root) {

18        

19         flag=true;

20         //判断是不是第一个被访问的节点

21         isFirstNode=true;

22  

23         inOrderTraversal(root);

24         return flag;

25     }

26    

27     void inOrderTraversal(TreeNode *root)

28     {

29         if(root==NULL)

30         {

31             return;

32         }

33        

34         if(!flag)

35         {

36             return;

37         }

38        

39         inOrderTraversal(root->left);

40        

41         //一旦发现不符合升序，则不是二叉排序树

42         if(!isFirstNode&&ret>=root->val)

43         {

44             flag=false;

45             return;

46         }

47  

48         ret=root->val;

49         isFirstNode=false;

50        

51         inOrderTraversal(root->right);

52     }

53    

54  

55 };