[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.

 

分析：二叉搜索树的中序遍历是从小到大排列的。

 

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
private:
  int pre;
  vector<int> vec;
public:
  Solution():pre(LONG_MIN) {}

  bool isValidBST(TreeNode* root)
  {
    vec.clear();
    if(root == NULL) return true;
    if(root->left==NULL && root->right==NULL) return true;
    isValid(root);
    
    for(int i=1; i<vec.size(); i++)
      if(vec[i-1] >= vec[i])
        return false;
     
       return true;
  }
  
  void isValid(TreeNode* root)
  {
    if(root)
    {
      isValid(root->left);
      vec.push_back(root->val);
      isValid(root->right);
    }
  }
};

 

方法二：设置上下界

class Solution
{
public:
  bool isValidBST(TreeNode* root)
  {
    if(root == NULL) return true;
    if(root->left==NULL && root->right==NULL) return true;
     
       return isValid(root, LONG_MIN, LONG_MAX);
  }
  
  bool isValid(TreeNode* root, int min, int max)
  {
    if(root==NULL) return true;
    if(root->val <= min || root->val >= max) return false;

    return isValid(root->left, min, root->val) && isValid(root->right, root->val, max);
  }
};